MathsCBSE Class 10 · Chapter 5

Arithmetic Progressions Class 10: Concepts, Formulas and Solved Questions

Updated 13 October 2026 · Fact-checked

Arithmetic Progressions is Class 10 Maths Chapter 5. An AP is a list of numbers where each term is obtained by adding the same fixed number, the common difference, to the previous term. The board asks you to find terms, the number of terms and sums using the nth term and sum formulas, and to solve daily-life word problems.

What Arithmetic Progressions is about

This chapter is about lists of numbers that grow or shrink by the same amount every step, such as $5, 9, 13, 17, \dots$ or $20, 17, 14, \dots$ . You learn two formulas, one for the $n$ th term and one for the sum of the first $n$ terms. You also learn why each one is true, because the syllabus asks you to derive them. With these two formulas you can jump to any term, count how many terms there are, or add hundreds of terms in one line.

The chapter leans on two earlier chapters. When you are given two conditions and must find the first term $a$ and common difference $d$ , you solve a pair of linear equations (Chapter 3). When you are given a sum and must find the number of terms $n$ , you usually get a quadratic equation (Chapter 4) and must reject the root that is not a natural number.

Board questions take a few clear shapes. Section A has MCQs and assertion-reason items that test $d$ , a particular term or a quick sum. Short answers ask you to find $k$ , find which term equals a number, or find the AP from two conditions. Long answers combine terms and sums, or ask for $n$ from a sum. Case studies give a real situation like savings, seating rows or production that rises by a fixed amount, and ask for a term and a total.

Weightage: This chapter is part of Unit II, Algebra, which carries 20 of the 80 theory marks together with Polynomials, Pair of Linear Equations in Two Variables and Quadratic Equations.Source: CBSE Mathematics syllabus 2026-27, Unit II Algebra

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5 marks per paper. See the questions

Key concepts

Arithmetic progression and common difference

NCERT section 5.2 Arithmetic Progressions

An arithmetic progression (AP) is a list of numbers in which every term after the first is got by adding the same fixed number to the term before it. That fixed number is the common difference $d$ . So $d = a_{2} - a_{1} = a_{3} - a_{2} = \dots$ . It can be positive, negative or zero.

To test whether a list is an AP, find the difference between every pair of consecutive terms. If all the differences are equal, it is an AP. Checking only the first pair is not enough.

For three terms $a$ , $b$ , $c$ in AP, $b - a = c - b$ , which gives $2 b = a + c$ . In words, the middle term is the average of its neighbours. Use this whenever the terms contain an unknown like $k$ .

Why study APs? Many real patterns change by a fixed amount each step: a salary with a fixed yearly increment, seats in the rows of a hall, fixed extra savings each month. The AP formulas let you reach any term or total without listing everything.

Worked example

For what value of $k$ are $2 k - 1$ , $3 k + 1$ and $5 k + 1$ three consecutive terms of an AP? Also write the three terms and the common difference.

Try it first, then show the solutionSolution

For consecutive terms of an AP, (second − first) = (third − second).
$(3 k + 1) - (2 k - 1) = (5 k + 1) - (3 k + 1)$
$k + 2 = 2 k$
$k = 2$
Terms: $2 (2) - 1 = 3$ , $3 (2) + 1 = 7$ , $5 (2) + 1 = 11$
Check: $7 - 3 = 4$ and $11 - 7 = 4$ , so $d = 4$ .

Answer: $k = 2$ . The terms are $3, 7, 11$ and $d = 4$ .

The nth term of an AP and its derivation

NCERT section 5.3 nth Term of an AP

Write the terms using $a$ and $d$ : $a_{1} = a$ , $a_{2} = a + d$ , $a_{3} = a + 2 d$ , $a_{4} = a + 3 d$ . Look at the pattern. The number of $d$ 's added is always one less than the position, because you start at the first term and take one step to reach each next term. So the $n$ th term is $a_{n} = a + (n - 1) d$ .

This single formula links four quantities: $a_{n}$ , $a$ , $n$ and $d$ . If you know any three, you can find the fourth.

To check whether a given number belongs to an AP, set $a_{n}$ equal to it and solve for $n$ . It is a term only if $n$ comes out as a natural number. A fraction or a negative value of $n$ means the number is not in the AP.

Notice also that $a_{n} = d n + (a - d)$ is linear in $n$ . The coefficient of $n$ is $d$ . So $a_{n} = 7 - 3 n$ describes an AP with $d = - 3$ .

Worked example

Which term of the AP $3, 8, 13, 18, \dots$ is 248? Is 301 a term of this AP? Give a reason.

Try it first, then show the solutionSolution

$a = 3$ , $d = 8 - 3 = 5$
Let $a_{n} = 248$ : $3 + (n - 1) 5 = 248$
$5 (n - 1) = 245 \Rightarrow n - 1 = 49 \Rightarrow n = 50$
Check: $a_{50} = 3 + 49 \times 5 = 248$ ✓
For 301: $3 + (n - 1) 5 = 301 \Rightarrow 5 (n - 1) = 298$
$n - 1 = \frac{298}{5} = 59.6$ , which is not a whole number, so $n$ is not a natural number.

Answer: 248 is the 50th term. 301 is not a term, because $n$ does not come out as a natural number.

Number of terms, nth term from the end and the middle term

NCERT section 5.3 nth Term of an AP

If an AP starts at $a$ and ends at $l$ , set $l = a + (n - 1) d$ to get $n = \frac{l - a}{d} + 1$ . The $+ 1$ matters. The gaps between terms number $\frac{l - a}{d}$ , and there is always one more term than gaps, just as a fence has one more post than gaps.

For the $n$ th term from the end, read the AP backwards. The first term becomes $l$ and the common difference becomes $- d$ . So the $n$ th term from the end is $l - (n - 1) d$ . Another way: if the AP has $m$ terms, the $n$ th from the end is the $(m - n + 1)$ th term from the start. Use one method to answer and the other to check.

For the middle term, first count the terms. If $m$ is odd, there is one middle term, the $(\frac{m + 1}{2})$ th. If $m$ is even, there are two middle terms, the $(\frac{m}{2})$ th and the $(\frac{m}{2} + 1)$ th.

Worked example

For the AP $4, 9, 14, \dots, 254$ , find (i) the number of terms, (ii) the 10th term from the end, and (iii) the middle term.

Try it first, then show the solutionSolution

$a = 4$ , $d = 5$ , $l = 254$
(i) $254 = 4 + (n - 1) 5 \Rightarrow n - 1 = 50 \Rightarrow n = 51$
(ii) 10th term from the end $= l - (10 - 1) d = 254 - 45 = 209$
Check: the 10th from the end is the $(51 - 10 + 1) = 42$ nd term, and $a_{42} = 4 + 41 \times 5 = 209$ ✓
(iii) 51 is odd, so the middle term is the $\frac{51 + 1}{2} = 26$ th term.
$a_{26} = 4 + 25 \times 5 = 129$

Answer: (i) 51 terms (ii) 209 (iii) 129

Finding the AP from given conditions

NCERT section 5.3 nth Term of an AP

An AP is fixed completely by two numbers, $a$ and $d$ . So when a question gives two facts about its terms, you get two equations in $a$ and $d$ . Write every term mentioned as $a + (n - 1) d$ , turn each sentence into an equation, and solve the pair as in Chapter 3.

A useful shortcut is $a_{m} - a_{k} = (m - k) d$ . For example, $a_{13} - a_{8} = 5 d$ , so the difference of two terms gives $d$ at once.

When a question talks about three numbers in AP with a given sum, call them $a - d$ , $a$ , $a + d$ . The $d$ 's cancel in the sum, so $3 a$ = sum gives $a$ straight away. For four numbers, use $a - 3 d$ , $a - d$ , $a + d$ , $a + 3 d$ . Here the common difference is $2 d$ .

Always finish by writing the first few terms and checking both given conditions.

Worked example

The 7th term of an AP is four times its 2nd term, and the 12th term is 2 more than three times its 4th term. Find the AP.

Try it first, then show the solutionSolution

$a_{7} = 4 a_{2}$ : $a + 6 d = 4 (a + d)$
$a + 6 d = 4 a + 4 d \Rightarrow 2 d = 3 a$ ...(1)
$a_{12} = 3 a_{4} + 2$ : $a + 11 d = 3 (a + 3 d) + 2$
$a + 11 d = 3 a + 9 d + 2 \Rightarrow 2 d = 2 a + 2 \Rightarrow d = a + 1$ ...(2)
Put (2) in (1): $2 (a + 1) = 3 a \Rightarrow a = 2$ , so $d = 3$
Check: $a_{7} = 20$ , $a_{2} = 5$ and $4 \times 5 = 20$ ✓. $a_{12} = 35$ , $a_{4} = 11$ and $3 \times 11 + 2 = 35$ ✓

Answer: The AP is $2, 5, 8, 11, \dots$ (with $a = 2$ , $d = 3$ ).

Sum of the first n terms and its derivation

NCERT section 5.4 Sum of First n Terms of an AP

Write the sum forwards and then backwards:

$S = a + (a + d) + \dots + [a + (n - 1) d]$

$S = [a + (n - 1) d] + [a + (n - 2) d] + \dots + a$

Add the two lines term by term. Every pair adds to the same value, $2 a + (n - 1) d$ , because as one term goes up by $d$ its partner goes down by $d$ . There are $n$ such pairs, so $2 S = n [2 a + (n - 1) d]$ . This gives $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

Since the last term is $l = a + (n - 1) d$ , the bracket equals $a + l$ . So $S_{n} = \frac{n}{2} (a + l)$ . In words: number of terms times the average of the first and last terms.

Use the first form when you know $a$ , $d$ and $n$ . Use the second when you know the last term. Taking $a = 1$ , $d = 1$ gives the sum of the first $n$ natural numbers, $\frac{n ( n + 1 )}{2}$ .

Worked example

Find the sum of all three-digit natural numbers that are divisible by 7.

Try it first, then show the solutionSolution

The smallest three-digit multiple of 7 is $105$ ( $7 \times 15$ ). The largest is $994$ ( $7 \times 142$ ).
The multiples form an AP with $a = 105$ , $d = 7$ , $l = 994$ .
$994 = 105 + (n - 1) 7 \Rightarrow n - 1 = \frac{889}{7} = 127 \Rightarrow n = 128$
$S_{128} = \frac{128}{2} (105 + 994) = 64 \times 1099$
$64 \times 1099 = 64 \times 1100 - 64 = 70400 - 64 = 70336$
Check: $n = 142 - 15 + 1 = 128$ ✓

Answer: The sum is $70336$ .

Finding the number of terms from a given sum

NCERT section 5.4 Sum of First n Terms of an AP

When $S_{n}$ is given and $n$ is unknown, substitute into $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . Because $n$ multiplies a bracket that itself contains $n$ , you get a **quadratic in $n$ **. Clear the fraction, write it in standard form and factorise (or use the quadratic formula).

The number of terms must be a natural number, so reject any root that is negative or a fraction, and say why.

Sometimes both roots are natural numbers. This happens when $d$ is negative. The terms become negative after some point, and the extra negative terms cancel some positive ones, so the same total appears twice. Both answers are then correct, and you should explain this in a line. Always check one value by direct substitution.

Worked example

How many terms of the AP $17, 15, 13, \dots$ must be taken so that their sum is 72? Explain why you get two answers.

Try it first, then show the solutionSolution

$a = 17$ , $d = - 2$
$S_{n} = \frac{n}{2} [34 + (n - 1) (- 2)] = \frac{n}{2} (36 - 2 n) = n (18 - n)$
$n (18 - n) = 72 \Rightarrow n^{2} - 18 n + 72 = 0$
$(n - 6) (n - 12) = 0 \Rightarrow n = 6$ or $n = 12$
Check: $17 + 15 + 13 + 11 + 9 + 7 = 72$ ✓
Terms 7 to 12 are $5, 3, 1, - 1, - 3, - 5$ , whose sum is $0$ , so adding them leaves the total at 72.

Answer: 6 terms or 12 terms. Both are correct, because the 7th to 12th terms add up to zero.

Finding the nth term when the sum of n terms is given

NCERT section 5.4 Sum of First n Terms of an AP

$S_{n}$ is the sum up to the $n$ th term and $S_{n - 1}$ is the sum up to the term before it. Subtracting removes everything except the last term. So $a_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$ , and the first term is simply $a_{1} = S_{1}$ .

When you find $S_{n - 1}$ , replace every $n$ by $(n - 1)$ and expand carefully with brackets. This is where most errors happen.

For an AP, $S_{n}$ always has the form $p n^{2} + q n$ , with no constant term. Then $a_{n} = 2 p n + (q - p)$ and $d = 2 p$ . So the coefficient of $n^{2}$ in $S_{n}$ is $\frac{d}{2}$ . Use this as a check. You can also find $d$ as $a_{2} - a_{1} = S_{2} - 2 S_{1}$ .

Check your $a_{n}$ by putting $n = 1$ : it must equal $S_{1}$ .

Worked example

The sum of the first $n$ terms of an AP is $S_{n} = 2 n^{2} + 3 n$ . Find the $n$ th term, the common difference and the 25th term.

Try it first, then show the solutionSolution

$S_{n - 1} = 2 (n - 1)^{2} + 3 (n - 1) = 2 n^{2} - 4 n + 2 + 3 n - 3 = 2 n^{2} - n - 1$
$a_{n} = S_{n} - S_{n - 1} = (2 n^{2} + 3 n) - (2 n^{2} - n - 1) = 4 n + 1$
Check: $a_{1} = 5$ and $S_{1} = 2 + 3 = 5$ ✓
$d$ = coefficient of $n$ in $a_{n}$ , so $d = 4$ (also $2 \times 2 = 4$ from $S_{n}$ ) ✓
$a_{25} = 4 (25) + 1 = 101$

Answer: $a_{n} = 4 n + 1$ , $d = 4$ , $a_{25} = 101$

Daily-life applications of AP

NCERT section 5.4 Sum of First n Terms of an AP

Word problems use the same two formulas. The skill is in modelling. First decide what changes by a fixed amount each step: rupees saved per week, logs per row, seats per row, distance per day. That fixed change is $d$ , and the first value is $a$ .

Next read the question carefully. Does it ask for one value, such as the amount saved in the 10th week or the seats in the last row? That is $a_{n}$ . Or does it ask for a total, such as total savings or total seats? That is $S_{n}$ . Mixing these up is the commonest error.

When the total is given and the time is asked, you get a quadratic in $n$ . Reject negative roots with a reason tied to the context, such as "weeks cannot be negative". Write the final answer as a sentence with units. Case-study questions in Section E are built exactly like this.

Worked example

Ravi saves ₹50 in the first week of the year and increases his weekly saving by ₹20 every week. In how many weeks will his total savings be ₹1,920? How much does he save in the last of these weeks?

Try it first, then show the solutionSolution

Weekly savings form an AP with $a = 50$ , $d = 20$ .
$S_{n} = \frac{n}{2} [100 + 20 (n - 1)] = \frac{n}{2} (20 n + 80) = 10 n^{2} + 40 n$
$10 n^{2} + 40 n = 1920 \Rightarrow n^{2} + 4 n - 192 = 0$
$(n + 16) (n - 12) = 0 \Rightarrow n = 12$ (reject $n = - 16$ , since the number of weeks cannot be negative)
Saving in the 12th week: $a_{12} = 50 + 11 \times 20 = 270$
Check: $S_{12} = \frac{12}{2} (50 + 270) = 6 \times 320 = 1920$ ✓

Answer: 12 weeks. He saves ₹270 in the 12th week.

Maths in real life: Arithmetic Progressions

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

The nth term of an AP and its derivation

Suppose a fresher gets two job offers. Offer A pays a starting monthly salary of ₹18,000 with a fixed yearly increment of ₹1,200 in the monthly salary. Offer B pays ₹21,000 a month to start, with a fixed yearly increment of ₹800. In each offer the monthly salary stays the same for all 12 months of a year, so the monthly salaries in successive years form an AP.

(i) From which year does the monthly salary under Offer A first exceed that under Offer B?

(ii) Find the total amount earned in the first 11 years under each offer. Which offer pays more over these 11 years?

(iii) After how many years will the total earnings under the two offers be equal?

Try it first, then show the solutionSolution

Offer A: monthly salary in year $n$ is $a_{n} = 18000 + (n - 1) 1200$ .
Offer B: monthly salary in year $n$ is $b_{n} = 21000 + (n - 1) 800$ .
(i) $a_{n} > b_{n}$ means $18000 + 1200 (n - 1) > 21000 + 800 (n - 1)$ , so $400 (n - 1) > 3000$ .
$n - 1 > 7.5$ , so the least natural number is $n = 9$ .
Check: year 8 gives A $= 18000 + 8400 = 26400$ and B $= 21000 + 5600 = 26600$ ; year 9 gives A $= 27600$ and B $= 27400$ .
(ii) Offer A, sum of the monthly salaries over 11 years: $\frac{11}{2} [36000 + 10 \times 1200] = \frac{11}{2} \times 48000 = 264000$ .
Each year has 12 months, so the total under A is $12 \times 264000 = 3168000$ , that is ₹31,68,000.
Offer B: $\frac{11}{2} [42000 + 10 \times 800] = \frac{11}{2} \times 50000 = 275000$ , so the total is $12 \times 275000 = 3300000$ , that is ₹33,00,000.
(iii) Equal totals: $\frac{n}{2} [36000 + 1200 (n - 1)] = \frac{n}{2} [42000 + 800 (n - 1)]$ .
Since $n \neq = 0$ : $400 (n - 1) = 6000$ , so $n - 1 = 15$ and $n = 16$ .
Check: A gives $8 [36000 + 18000] = 432000$ and B gives $8 [42000 + 12000] = 432000$ as sums of monthly salaries. They match.

Answer: (i) From the 9th year (₹27,600 against ₹27,400). (ii) Offer A: ₹31,68,000; Offer B: ₹33,00,000, so Offer B pays more over the first 11 years. (iii) After 16 years the total earnings are equal.

Memory hook: Salary ladder: year $n$ has $(n - 1)$ increments added, so a bigger step can overtake a higher start.

Number of terms, nth term from the end and the middle term

Imagine a long street in your town where the houses on one side are numbered 7, 11, 15, 19, and so on, up to 203. The numbers go up by 4 from one house to the next.

(i) How many houses are on this side of the street?

(ii) Find the number of the 10th house counted from the end of the street.

(iii) Find the sum of the house numbers from the 11th house to the 30th house.

Try it first, then show the solutionSolution

Here $a = 7$ , $d = 4$ and the last term is $l = 203$ .
(i) $n = \frac{l - a}{d} + 1 = \frac{203 - 7}{4} + 1$ .
$n = \frac{196}{4} + 1 = 49 + 1 = 50$ houses.
(ii) The 10th term from the end is $l - (n - 1) d = 203 - 9 \times 4 = 203 - 36 = 167$ .
Check: the 10th from the end is the $(50 - 10 + 1)$ th, that is the 41st, term from the start, and $a_{41} = 7 + 40 \times 4 = 167$ . Both methods agree.
(iii) $a_{11} = 7 + 10 \times 4 = 47$ and $a_{30} = 7 + 29 \times 4 = 123$ .
The 11th to 30th houses make $30 - 11 + 1 = 20$ terms of an AP.
Sum $= \frac{20}{2} (47 + 123) = 10 \times 170 = 1700$ .
Check: $S_{30} - S_{10} = \frac{30}{2} (14 + 29 \times 4) - \frac{10}{2} (14 + 9 \times 4) = 15 \times 130 - 5 \times 50 = 1950 - 250 = 1700$ .

Answer: (i) 50 houses. (ii) The 10th house from the end is number 167. (iii) The house numbers from the 11th to the 30th house add up to 1700.

Memory hook: Street numbers: count the gaps and add one more house, and a block of houses is just $S_{30} - S_{10}$ .

Finding the AP from given conditions

Suppose you take a weekly maths test marked out of 100, and your marks go up by the same amount each week. You scored 62 in the 4th test and 82 in the 9th test.

(i) Find the marks in your first test and the weekly increase.

(ii) Marks cannot go above 100. For how many tests can this pattern continue, and what are your total marks over those tests?

(iii) In which test does your running total of marks first exceed 700?

Try it first, then show the solutionSolution

Let the first test's marks be $a$ and the weekly increase be $d$ .
4th test: $a + 3 d = 62$ ... (1); 9th test: $a + 8 d = 82$ ... (2)
Subtract (1) from (2): $5 d = 20$ , so $d = 4$ , and then $a = 62 - 12 = 50$ .
(ii) We need $50 + (n - 1) 4 \leq 100$ , so $(n - 1) 4 \leq 50$ and $n - 1 \leq 12.5$ .
So $n \leq 13$ : the 13th test gives $50 + 48 = 98$ , while the 14th would need 102, which is impossible.
Total over 13 tests: $S_{13} = \frac{13}{2} (50 + 98) = 13 \times 74 = 962$ .
(iii) $S_{n} = \frac{n}{2} [100 + 4 (n - 1)] = 2 n^{2} + 48 n$ .
$S_{10} = 200 + 480 = 680$ , which does not exceed 700.
$S_{11} = 242 + 528 = 770$ , which exceeds 700. Every score is positive, so the running total only grows.

Answer: (i) You scored 50 in the first test, and your marks rose by 4 each week. (ii) The pattern can continue for 13 tests, with a total of 962 marks. (iii) In the 11th test, when the running total reaches 770.

Memory hook: Two test scores fix the whole AP: their difference gives $d$ , then $d$ gives $a$ , and the 100-mark ceiling caps $n$ .

Sum of the first n terms and its derivation

Imagine your family is setting up a triangular display of diyas for Diwali. The top row has 3 diyas, and each row below has 2 more diyas than the row above it. You have 255 diyas and want to use all of them, and suppose each extra diya costs ₹4.

(i) How many rows will the display have, and how many diyas are in the bottom row?

(ii) Your aunt wants the bottom row to have at least 40 diyas. What is the least number of rows needed now, how many extra diyas must be bought, and what will they cost?

Try it first, then show the solutionSolution

The rows form an AP with $a = 3$ and $d = 2$ .
(i) $S_{n} = \frac{n}{2} [6 + 2 (n - 1)] = \frac{n}{2} (2 n + 4) = n (n + 2)$ .
$n (n + 2) = 255$ gives $n^{2} + 2 n - 255 = 0$ , that is $(n + 17) (n - 15) = 0$ .
Reject $n = - 17$ , since the number of rows cannot be negative. So $n = 15$ .
Bottom row: $a_{15} = 3 + 14 \times 2 = 31$ diyas. Check: $\frac{15}{2} (3 + 31) = 15 \times 17 = 255$ .
(ii) We need $a_{n} = 3 + 2 (n - 1) = 2 n + 1 \geq 40$ , so $n \geq 19.5$ , and the least natural number is $n = 20$ .
The 20th row has $2 (20) + 1 = 41$ diyas; a 19th bottom row would have only 39.
Total for 20 rows: $S_{20} = 20 \times 22 = 440$ diyas.
Extra diyas $= 440 - 255 = 185$ , and their cost is $185 \times 4 = 740$ , that is ₹740.

Answer: (i) 15 rows, with 31 diyas in the bottom row. (ii) 20 rows, with 41 diyas in the bottom row, so 185 extra diyas are needed, costing ₹740.

Memory hook: Diya triangle: pair the top row with the bottom row; every pair has the same total, so the sum is (rows) times (average of first and last rows).

Finding the number of terms from a given sum

Suppose you start saving in a piggy bank for a school trip. You put in ₹20 in the first week, and each week you put in ₹5 more than the week before. The trip will cost you ₹1,350.

After how many weeks will your total savings be exactly ₹1,350? How much will you put in during the last week?

Try it first, then show the solutionSolution

The weekly savings form an AP with $a = 20$ and $d = 5$ . We need $S_{n} = 1350$ .
$\frac{n}{2} [2 \times 20 + (n - 1) 5] = 1350$ .
$\frac{n}{2} [40 + 5 n - 5] = 1350$ , so $n (5 n + 35) = 2700$ .
$5 n^{2} + 35 n - 2700 = 0$ . Divide by 5: $n^{2} + 7 n - 540 = 0$ .
Factorise: $(n + 27) (n - 20) = 0$ , so $n = 20$ or $n = - 27$ .
Reject $n = - 27$ , because the number of weeks cannot be negative. So $n = 20$ .
Last week: $a_{20} = 20 + 19 \times 5 = 20 + 95 = 115$ .
Check: $S_{20} = \frac{20}{2} (20 + 115) = 10 \times 135 = 1350$ . Correct.

Answer: Your savings reach exactly ₹1,350 after 20 weeks. You put in ₹115 in the 20th week.

Memory hook: Piggy bank: total given and weeks asked means a quadratic in $n$ . Throw out the negative week.

Finding the nth term when the sum of n terms is given

Suppose you are reading a long novel during the summer holidays. You notice that the total number of pages you have read by the end of day $n$ is always $S_{n} = 2 n^{2} + 18 n$ .

How many pages do you read on the 10th day? On which day do you read exactly 60 pages?

Try it first, then show the solutionSolution

Pages read on day $n$ : $a_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$ .
$S_{n - 1} = 2 (n - 1)^{2} + 18 (n - 1) = 2 n^{2} - 4 n + 2 + 18 n - 18 = 2 n^{2} + 14 n - 16$ .
$a_{n} = (2 n^{2} + 18 n) - (2 n^{2} + 14 n - 16) = 4 n + 16$ .
Check for day 1: $a_{1} = 4 + 16 = 20$ , and $S_{1} = 2 + 18 = 20$ . They match, so $a_{n} = 4 n + 16$ holds for every day.
Day 10: $a_{10} = 4 \times 10 + 16 = 56$ pages.
Check: $S_{10} - S_{9} = (200 + 180) - (162 + 162) = 380 - 324 = 56$ . Correct.
For 60 pages: $4 n + 16 = 60$ , so $4 n = 44$ and $n = 11$ .

Answer: You read 56 pages on the 10th day, and exactly 60 pages on the 11th day.

Memory hook: Reading log: today's pages = total so far minus total till yesterday, that is $S_{n} - S_{n - 1}$ .

Topics in this chapter

Arithmetic Progressions formula sheet

Common difference: $d = a_{2} - a_{1} = a_{k + 1} - a_{k}$; Always subtract a term from the term after it (later minus earlier). Check at least two pairs before calling a list an AP.
General form of an AP: $a, a + d, a + 2 d, a + 3 d, \dots$; Write terms this way when you build equations from given conditions.
nth term: $a_{n} = a + (n - 1) d$; Use it to find any term, to find which term equals a number ( $n$ must be a natural number), or to form equations in $a$ and $d$ .
Difference of two terms: $a_{m} - a_{k} = (m - k) d$; A quick way to get $d$ when two terms are given, and for questions like 'which term is 84 more than the 20th term'.
nth term from the end: $nth term from end = l - (n - 1) d$; $l$ is the last term. This is the same as the $(m - n + 1)$ th term from the start when there are $m$ terms.
Number of terms from a to l: $n = \frac{l - a}{d} + 1$; Use it for counting multiples in a range. Never forget the $+ 1$ .
Condition for three terms in AP: $2 b = a + c$; If $a$ , $b$ , $c$ are consecutive terms of an AP. Use it to find an unknown $k$ .
Choosing unknown terms: $a - d, a, a + d; a - 3 d, a - d, a + d, a + 3 d$; Use these when the sum of three or four terms in AP is given, so that $d$ cancels. For the four-term set, the common difference is $2 d$ .
Middle term: $odd m : (\frac{m + 1}{2}) th term; even m : (\frac{m}{2}) th and (\frac{m}{2} + 1) th terms$; Here $m$ is the total number of terms. Count the terms first.
Sum of first n terms: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$; Use it when $a$ , $d$ and $n$ are known, or to form a quadratic in $n$ when the sum is given.
Sum using the last term: $S_{n} = \frac{n}{2} (a + l)$; Faster when the first and last terms are known, as in sums of multiples.
nth term from the sum: $a_{n} = S_{n} - S_{n - 1} (n \geq 2), a_{1} = S_{1}$; Use it when $S_{n}$ is given as an expression in $n$ .
Sum of first n natural numbers: $1 + 2 + 3 + \dots + n = \frac{n ( n + 1 )}{2}$; A special case of the sum formula with $a = 1$ , $d = 1$ .
Shape of a_n and S_n: $a_{n} = d n + (a - d), S_{n} = \frac{d}{2} n^{2} + (a - \frac{d}{2}) n$; $a_{n}$ is linear in $n$ and $S_{n}$ is quadratic with no constant term. Use this to read off $d$ quickly or to check an answer.

How the board asks Arithmetic Progressions

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Find k so that the given terms are in AP	Three expressions in $k$ (sometimes squares or fractions) are said to be consecutive terms of an AP. Find $k$ and sometimes the terms. It often appears as an MCQ or a 2-mark question.	Use $2 \times$ (middle) $=$ first $+$ third. Solve for $k$ , then substitute back and check that the two differences are equal.
Find a term, or test whether a number is a term	Find the 30th term, find which term equals a given number, find which term is a given amount more than another term, or decide whether a number belongs to the AP.	Find $a$ and $d$ first. Set $a + (n - 1) d$ equal to the number and solve. State clearly that $n$ is or is not a natural number. For 'more than' questions use $a_{m} - a_{k} = (m - k) d$ .
Count terms, nth term from the end, middle term	An AP with a last term is given. Find how many terms it has, the 8th term from the end, or its middle term(s). This also covers counting multiples of a number in a range.	Find $n$ from $l = a + (n - 1) d$ . Use $l - (n - 1) d$ for terms from the end. For the middle term, decide whether $n$ is odd or even first.
Find the AP from two conditions	Two facts are given, such as 'the 4th term is 11 and the sum of the 5th and 7th terms is 34', or 'three numbers in AP have sum 27 and product 648'. You must find the AP.	Turn each fact into an equation in $a$ and $d$ and solve the pair. For three numbers in AP, take $a - d$ , $a$ , $a + d$ . Write the AP and check both conditions.
Find a sum, including sums of multiples	Find the sum of the first 25 terms, the sum of all two-digit odd numbers, or the sum of the terms from the 11th to the 30th.	Identify $a$ , $d$ , $n$ (and $l$ if known), then apply the sum formula. For a block of terms in the middle, use $S_{30} - S_{10}$ , or treat the block as a new AP starting at $a_{11}$ .
Find n when the sum is given	How many terms must be taken to get a given sum? It often comes with a negative $d$ , which leads to two valid answers.	Substitute into $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , form the quadratic and factorise. Reject non-natural roots with a reason. If both roots are natural numbers, explain why both work.
S_n given as an expression in n	$S_{n}$ is given, for example as $3 n^{2} - 2 n$ . Find the first term, $d$ , the $n$ th term or a particular term. It also appears as an assertion-reason item.	$a_{1} = S_{1}$ and $a_{n} = S_{n} - S_{n - 1}$ . Expand $S_{n - 1}$ with full brackets. Check that the coefficient of $n$ in $a_{n}$ equals twice the coefficient of $n^{2}$ in $S_{n}$ .
Word problems and case studies	Savings, instalments, salary increments, seats in rows, logs in a stack, plants in a garden or production that rises by a fixed amount. Section E asks for a term (1 mark), a value (1 mark) and a total or the number of steps (2 marks).	Define $a$ and $d$ in words with units. Decide whether each part asks for a single value ( $a_{n}$ ) or a total ( $S_{n}$ ). Solve, reject impossible roots using the context, and answer in a sentence with units.

Tricks to remember Arithmetic Progressions

Short memory aids. Say each one aloud once, then use it on a question the same day.

n − 1 steps
To reach the $n$ th term you take $n - 1$ steps from the first term, each of size $d$ . So $a_{n} = a + (n - 1) d$ , never $a + n d$ .
Use it when: Say it before every $n$ th-term calculation.
Fence posts
Gaps $= \frac{l - a}{d}$ and terms $=$ gaps $+ 1$ , just as a fence has one more post than gaps.
Use it when: Counting multiples in a range, or the number of terms in a finite AP.
Flip it for the end
Reading an AP backwards turns $a$ into $l$ and $d$ into $- d$ . So the $n$ th term from the end is $l - (n - 1) d$ .
Use it when: Questions on terms from the end, and checking with the $(m - n + 1)$ th term from the start.
The S₁ test
Whatever $a_{n}$ you get from $S_{n}$ , put $n = 1$ . It must equal $S_{1}$ . Also, for an AP the coefficient of $n^{2}$ in $S_{n}$ is half of $d$ .
Use it when: A ten-second check on any $S_{n}$ to $a_{n}$ question.
Pair-up picture for S_n
First plus last, second plus second-last: every pair is equal. So the sum is (number of terms) $\times$ (average of first and last) $= \frac{n}{2} (a + l)$ .
Use it when: Remembering and deriving the sum formula in the exam.
Later minus earlier
$d$ is always the later term minus the earlier term. A falling AP has a negative $d$ .
Use it when: Avoiding sign errors with APs like $17, 15, 13, \dots$

Arithmetic Progressions: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: 10th term from a sum formula · 1 mark · MCQBoard level

The sum of the first $n$ terms of an AP is $3 n^{2} - n$ . The 10th term of the AP is
(a) 56
(b) 290
(c) 59
(d) 53

Try it first, then show the solutionSolution

$a_{10} = S_{10} - S_{9}$ , because the 10th term is what gets added to the first 9 terms to make the first 10.
$S_{10} = 3 (100) - 10 = 290$ and $S_{9} = 3 (81) - 9 = 234$ .
$a_{10} = 290 - 234 = 56$ .
Check: $a_{1} = S_{1} = 2$ and $a_{2} = S_{2} - S_{1} = 10 - 2 = 8$ , so $d = 6$ and $a_{10} = 2 + 9 \times 6 = 56$ .
Option (b) is the trap: 290 is $S_{10}$ , not $a_{10}$ .

Answer: (a) 56

Example 2: Can 1000 be a term? · 1 mark · Assertion-reasonBoard level

Assertion (A): 1000 is not a term of the AP $3, 7, 11, 15, \dots$
Reason (R): A number $x$ is a term of an AP with first term $a$ and common difference $d$ only if $\frac{x - a}{d}$ is a whole number.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Check R: $x = a + (n - 1) d$ gives $n - 1 = \frac{x - a}{d}$ , and $n - 1$ must be 0, 1, 2, ... So R is true.
Check A: here $a = 3$ , $d = 4$ , so $\frac{1000 - 3}{4} = \frac{997}{4}$ , which is not a whole number.
So 1000 is not a term, and A is true.
We used R directly to decide A, so R explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: When two APs meet · 3 marks · Short answerBoard level

For what value of $n$ are the $n$ th terms of the APs $7, 12, 17, \dots$ and $25, 28, 31, \dots$ equal? Find that common term.

Try it first, then show the solutionSolution

First AP: $a = 7$ , $d = 5$ , so $a_{n} = 7 + (n - 1) 5 = 5 n + 2$ . [½ mark]
Second AP: $a = 25$ , $d = 3$ , so $b_{n} = 25 + (n - 1) 3 = 3 n + 22$ . [½ mark]
Equate: $5 n + 2 = 3 n + 22$ , so $2 n = 20$ and $n = 10$ . [1 mark]
Common term: $5 (10) + 2 = 52$ . [½ mark]
Check in the second AP: $3 (10) + 22 = 52$ . [½ mark]

Answer: The 10th terms are equal, and the common term is 52.

Example 4: Sum of numbers with a given remainder · 3 marks · Short answerBoard level

Find the sum of all two-digit natural numbers that leave a remainder 3 when divided by 5.

Try it first, then show the solutionSolution

Such numbers are of the form $5 k + 3$ . The smallest two-digit one is 13 and the largest is 98, since $98 = 5 \times 19 + 3$ . They form the AP $13, 18, 23, \dots, 98$ with $d = 5$ . [½ mark]
Number of terms: $98 = 13 + (n - 1) 5$ , so $(n - 1) 5 = 85$ , $n - 1 = 17$ , $n = 18$ . [1 mark]
Sum: $S_{18} = \frac{18}{2} (13 + 98) = 9 \times 111 = 999$ . [1 mark]
Check with the other formula: $S_{18} = 9 [26 + 17 \times 5] = 9 \times 111 = 999$ . [½ mark]

Answer: The sum is 999.

Example 5: Derive the sum formula and use it · 5 marks · Long answerHarder board level

Derive the formula for the sum of the first $n$ terms of an AP with first term $a$ and common difference $d$ . Hence find the sum of the first 25 terms of the AP whose $n$ th term is $7 - 3 n$ .

Try it first, then show the solutionSolution

Write the sum forwards: $S = a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d]$ . [½ mark]
Write it backwards: $S = [a + (n - 1) d] + [a + (n - 2) d] + \dots + a$ . [½ mark]
Add the two lines term by term. Each pair adds to $2 a + (n - 1) d$ , and there are $n$ pairs, so $2 S = n [2 a + (n - 1) d]$ . [1 mark]
Hence $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . [½ mark]
Since the last term is $l = a + (n - 1) d$ , this is also $S_{n} = \frac{n}{2} (a + l)$ . [½ mark]
For $a_{n} = 7 - 3 n$ : $a_{1} = 7 - 3 = 4$ and $a_{25} = 7 - 75 = - 68$ . [1 mark]
$S_{25} = \frac{25}{2} (4 + (- 68)) = \frac{25}{2} \times (- 64) = - 800$ . [1 mark]

Answer: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , and the sum of the first 25 terms is $- 800$ .

Example 6: Four terms from a sum and a ratio · 5 marks · Long answerHarder board level

The sum of four consecutive terms of an AP is 32, and the ratio of the product of the first and last terms to the product of the two middle terms is $7 : 15$ . Find the four terms.

Try it first, then show the solutionSolution

Take the terms as $a - 3 d$ , $a - d$ , $a + d$ , $a + 3 d$ . They are in AP with common difference $2 d$ , and the $d$ 's cancel when you add. [1 mark]
Sum: $4 a = 32$ , so $a = 8$ . [1 mark]
Ratio: $\frac{( 8 - 3 d ) ( 8 + 3 d )}{( 8 - d ) ( 8 + d )} = \frac{7}{15}$ , that is $\frac{64 - 9 d ^{2}}{64 - d ^{2}} = \frac{7}{15}$ . [1 mark]
Cross-multiply: $960 - 135 d^{2} = 448 - 7 d^{2}$ , so $128 d^{2} = 512$ , $d^{2} = 4$ , $d = \pm 2$ . [1 mark]
With $d = 2$ the terms are 2, 6, 10, 14; with $d = - 2$ they are 14, 10, 6, 2. Check: $2 \times 14 = 28$ , $6 \times 10 = 60$ , and $\frac{28}{60} = \frac{7}{15}$ . [1 mark]

Answer: The terms are 2, 6, 10, 14 (or the same numbers in the order 14, 10, 6, 2).

Example 7: Ratio of terms from a ratio of sums · 3 marks · Short answerHarder board level

The sums of the first $n$ terms of two APs are in the ratio $(3 n + 8) : (7 n + 15)$ . Find the ratio of their 12th terms.

Try it first, then show the solutionSolution

Let the APs have first terms $a$ , $A$ and common differences $d$ , $D$ . Then $\frac{S _{n}}{S _{n}^{'}} = \frac{\frac{n}{2} [ 2 a + ( n - 1 ) d ]}{\frac{n}{2} [ 2 A + ( n - 1 ) D ]} = \frac{2 a + ( n - 1 ) d}{2 A + ( n - 1 ) D} = \frac{3 n + 8}{7 n + 15}$ . [1 mark]
Divide the top and bottom of the middle fraction by 2: $\frac{a + \frac{n - 1}{2} d}{A + \frac{n - 1}{2} D} = \frac{3 n + 8}{7 n + 15}$ . [½ mark]
The 12th terms are $a + 11 d$ and $A + 11 D$ , so we need $\frac{n - 1}{2} = 11$ , that is $n = 23$ . [½ mark]
Put $n = 23$ : $\frac{a + 11 d}{A + 11 D} = \frac{3 ( 23 ) + 8}{7 ( 23 ) + 15} = \frac{77}{176} = \frac{7}{16}$ . [½ mark]
Check with sums that fit the ratio, $S_{n} = n (3 n + 8)$ and $S_{n}^{'} = n (7 n + 15)$ : their $n$ th terms are $6 n + 5$ and $14 n + 8$ , giving 12th terms 77 and 176, and $\frac{77}{176} = \frac{7}{16}$ . [½ mark]

Answer: The ratio of the 12th terms is $7 : 16$ .

Example 8: Divisible by 9 but not by 6 · 3 marks · Short answerHarder board level

Find the sum of all integers between 100 and 500 that are divisible by 9 but not by 6.

Try it first, then show the solutionSolution

Multiples of 9 between 100 and 500: $108, 117, \dots, 495$ , since $108 = 9 \times 12$ and $495 = 9 \times 55$ . Number of terms $= 55 - 12 + 1 = 44$ , so their sum is $\frac{44}{2} (108 + 495) = 22 \times 603 = 13266$ . [1 mark]
A number divisible by both 9 and 6 is divisible by their LCM, 18. Multiples of 18 in the range run from $108 = 18 \times 6$ to $486 = 18 \times 27$ , which is 22 terms with sum $\frac{22}{2} (108 + 486) = 11 \times 594 = 6534$ . [1 mark]
Required sum $= 13266 - 6534 = 6732$ . [½ mark]
Check: the numbers wanted are the odd multiples of 9, $117, 135, \dots, 495$ , an AP with $d = 18$ and $\frac{495 - 117}{18} + 1 = 22$ terms, so the sum is $11 (117 + 495) = 11 \times 612 = 6732$ . [½ mark]

Answer: The sum is 6732.

Example 9: From two conditions to the least n · 5 marks · Long answerHarder board level

The sum of the 4th and 8th terms of an AP is 24, and the sum of its 6th and 10th terms is 44. Find the AP and the sum of its first 10 terms. Also find the least number of terms whose sum exceeds 500.

Try it first, then show the solutionSolution

$a_{4} + a_{8} = (a + 3 d) + (a + 7 d) = 24$ , so $a + 5 d = 12$ ...(1) [½ mark]
$a_{6} + a_{10} = (a + 5 d) + (a + 9 d) = 44$ , so $a + 7 d = 22$ ...(2) [½ mark]
Subtract (1) from (2): $2 d = 10$ , so $d = 5$ and $a = 12 - 25 = - 13$ . The AP is $- 13, - 8, - 3, 2, 7, \dots$ [1 mark]
$S_{10} = \frac{10}{2} [2 (- 13) + 9 \times 5] = 5 \times 19 = 95$ . [1 mark]
$S_{n} = \frac{n}{2} [- 26 + 5 (n - 1)] = \frac{n ( 5 n - 31 )}{2}$ . From the 4th term on every term is positive, so $S_{n}$ keeps increasing after $n = 3$ . [½ mark]
$S_{17} = \frac{17 \times 54}{2} = 459$ , which is less than 500. [½ mark]
$S_{18} = \frac{18 \times 59}{2} = 531$ , which exceeds 500. So the least number of terms is 18. [1 mark]

Answer: The AP is $- 13, - 8, - 3, 2, \dots$ and $S_{10} = 95$ . The least number of terms whose sum exceeds 500 is 18.

Example 10: Find the error: nth term from a sum · 3 marks · Short answerHOTS / competency

The sum of the first $n$ terms of an AP is $S_{n} = 4 n^{2} + n$ . A student writes: "The nth term is $\frac{S _{n}}{n} = 4 n + 1$ , so the 10th term is 41." Find the mistake, and find the correct $n$ th term and 10th term.

Try it first, then show the solutionSolution

The mistake: $\frac{S _{n}}{n}$ is the average of the first $n$ terms, not the $n$ th term. For $n = 1$ it happens to agree, which hides the error. [1 mark]
Correct method: $a_{n} = S_{n} - S_{n - 1} = (4 n^{2} + n) - [4 (n - 1)^{2} + (n - 1)] = 4 n^{2} + n - 4 n^{2} + 8 n - 4 - n + 1 = 8 n - 3$ . [1 mark]
So $a_{10} = 80 - 3 = 77$ . Check: $S_{1} = 5 = a_{1}$ and $S_{2} - S_{1} = 18 - 5 = 13 = a_{2}$ , both fitting $8 n - 3$ . [1 mark]

Answer: The $n$ th term is $8 n - 3$ , not $4 n + 1$ , and the 10th term is 77, not 41.

Example 11: Seats in an auditorium · 5 marks · Long answerHOTS / competency

An auditorium has 20 seats in the first row, and each row after that has 2 more seats than the row in front of it. The auditorium has 1,100 seats in all. (i) How many rows are there? (ii) How many seats are in the last row? (iii) Tickets in the first 10 rows cost ₹300 each and all other tickets cost ₹200 each. Find the money collected when every seat is sold.

Try it first, then show the solutionSolution

Seats per row form an AP with $a = 20$ , $d = 2$ . So $S_{n} = \frac{n}{2} [40 + 2 (n - 1)] = n (n + 19)$ . [1 mark]
$n (n + 19) = 1100$ gives $n^{2} + 19 n - 1100 = 0$ , that is $(n + 44) (n - 25) = 0$ . [1 mark]
$n = 25$ , since the number of rows cannot be negative, so $n = - 44$ is rejected. [1 mark]
Last row: $a_{25} = 20 + 24 \times 2 = 68$ seats. Check: $\frac{25}{2} (20 + 68) = 25 \times 44 = 1100$ . [1 mark]
First 10 rows: $S_{10} = 10 \times 29 = 290$ seats, leaving $1100 - 290 = 810$ seats. Money $= 290 \times 300 + 810 \times 200 = 87000 + 162000 = 249000$ . [1 mark]

Answer: (i) 25 rows (ii) 68 seats (iii) ₹2,49,000

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Green road plantation drive

A village panchayat in Madhya Pradesh runs a plantation drive along a new district road. Volunteers plant 40 saplings on the first day. Every day after that they plant 6 more saplings than on the day before, so the daily numbers are 40, 46, 52, and so on. The panchayat keeps a daily register and wants to plan how many saplings to order from the forest nursery. Saplings cost ₹15 each, and the drive continues until the target number of saplings is reached.

(i) · 1 mark
How many saplings are planted on the 15th day?
Answer
Answer: 124 saplings
$a = 40$ , $d = 6$ . $a_{15} = 40 + 14 \times 6 = 40 + 84 = 124$ .
(ii) · 1 mark
On which day are exactly 100 saplings planted?
Answer
Answer: The 11th day
$40 + (n - 1) 6 = 100$ gives $(n - 1) 6 = 60$ , so $n - 1 = 10$ and $n = 11$ .
(iii) · 2 marks
The target is 1,360 saplings in all. In how many days will the target be reached, and what is the cost of the saplings?
Answer
Answer: 16 days; ₹20,400
$S_{n} = \frac{n}{2} [80 + 6 (n - 1)] = n (3 n + 37) = 1360$ , so $3 n^{2} + 37 n - 1360 = 0$ , that is $(3 n + 85) (n - 16) = 0$ . Reject the negative root, so $n = 16$ . Check: $S_{16} = 8 (80 + 90) = 1360$ . Cost $= 1360 \times 15 =$ ₹20,400.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Which term of the AP $7, 11, 15, \dots$ is 84 more than its 20th term? (2 marks)

Step	Marks
Identify $a = 7$ , $d = 4$ and find $a_{20} = 7 + 19 \times 4 = 83$	0.5
The required term is $83 + 84 = 167$	0.5
$7 + (n - 1) 4 = 167 \Rightarrow 4 (n - 1) = 160 \Rightarrow n - 1 = 40$	0.5
$n = 41$ , so the 41st term is 84 more than the 20th term	0.5

The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28th term of the AP. (3 marks)

Step	Marks
$S_{7} = \frac{7}{2} (2 a + 6 d) = 63 \Rightarrow a + 3 d = 9$	1
The sum of the first 14 terms is $S_{14} = 63 + 161 = 224$	0.5
$\frac{14}{2} (2 a + 13 d) = 224 \Rightarrow 2 a + 13 d = 32$	0.5
From $2 a + 6 d = 18$ , subtracting gives $7 d = 14$ , so $d = 2$ and $a = 3$	0.5
$a_{28} = 3 + 27 \times 2 = 57$	0.5

The 4th term of an AP is 11 and the sum of its 5th and 7th terms is 34. Find the first term and the common difference. Also find the sum of the first 20 terms, and how many terms of the AP must be taken to get a sum of 155. (5 marks)

Step	Marks
$a_{4} = a + 3 d = 11$	0.5
$a_{5} + a_{7} = (a + 4 d) + (a + 6 d) = 34 \Rightarrow a + 5 d = 17$	0.5
Subtracting: $2 d = 6$ , so $d = 3$ and $a = 2$	1
$S_{20} = \frac{20}{2} [2 (2) + 19 (3)] = 10 \times 61 = 610$	1
$\frac{n}{2} [4 + 3 (n - 1)] = 155 \Rightarrow n (3 n + 1) = 310 \Rightarrow 3 n^{2} + n - 310 = 0$	1
$(n - 10) (3 n + 31) = 0$ , so $n = 10$ . $n = - \frac{31}{3}$ is rejected because $n$ must be a natural number.	1

Where students lose marks in Arithmetic Progressions

The slips examiners see most often, and what to write instead.

Mistake: Using $a + n d$ instead of $a + (n - 1) d$
Students think 'the 20th term needs 20 jumps' and forget that the first term already has no $d$ added.
Wrong working
For $3, 8, 13, \dots$ : $a_{20} = 3 + 20 \times 5 = 103$
Right working
$a_{20} = 3 + (20 - 1) \times 5 = 3 + 95 = 98$
Do this: Write the formula $a_{n} = a + (n - 1) d$ as the first line every time, then substitute.
Usually costs: The answer mark, plus every later step that uses the wrong term.
Mistake: Wrong sign of $d$ for a decreasing AP
Students subtract the smaller number from the bigger out of habit.
Wrong working
$17, 15, 13, \dots$ : $d = 17 - 15 = 2$ , so $a_{10} = 17 + 9 (2) = 35$
Right working
$d = 15 - 17 = - 2$ , so $a_{10} = 17 + 9 (- 2) = - 1$
Do this: Always compute later term minus earlier term. If the terms are falling, expect a negative $d$ .
Usually costs: Usually all the marks after the first step, since everything depends on $d$.
Mistake: Forgetting the $+ 1$ when counting terms
Dividing the range by $d$ counts the gaps, not the terms.
Wrong working
Three-digit multiples of 7: $n = \frac{994 - 105}{7} = 127$
Right working
$n = \frac{994 - 105}{7} + 1 = 128$ , or solve $994 = 105 + (n - 1) 7$
Do this: Find $n$ from $l = a + (n - 1) d$ rather than a shortcut, or check with a small case like $7, 14, 21$ .
Usually costs: ½ to 1 mark for $n$, and the sum then comes out wrong as well.
Mistake: Not rejecting an impossible value of $n$
Students solve the quadratic and write both roots as the answer.
Wrong working
$n^{2} + 4 n - 192 = 0 \Rightarrow n = 12$ or $n = - 16$ . So the answer is 12 or −16 weeks.
Right working
$(n + 16) (n - 12) = 0$ , so $n = 12$ . $n = - 16$ is rejected because the number of weeks must be a natural number.
Do this: After solving, write one line rejecting each root that is negative or a fraction, with the reason.
Usually costs: ½ to 1 mark for the final answer.
Mistake: Errors in finding $a_{n}$ from $S_{n}$
Students take $a_{n} = S_{n}$ , or replace only one $n$ by $(n - 1)$ , or drop brackets while expanding.
Wrong working
$S_{n} = 2 n^{2} + 3 n$ , so $S_{n - 1} = 2 n^{2} + 3 n - 1$ and $a_{n} = 1$
Right working
$S_{n - 1} = 2 (n - 1)^{2} + 3 (n - 1) = 2 n^{2} - n - 1$ , so $a_{n} = (2 n^{2} + 3 n) - (2 n^{2} - n - 1) = 4 n + 1$
Do this: Replace every $n$ with $(n - 1)$ inside brackets, expand slowly, then check $a_{1} = S_{1}$ .
Usually costs: Most of a 2- or 3-mark question.
Mistake: Confusing a single term with the total
In word problems, both 'how much in the 12th week' and 'how much in 12 weeks' sound similar.
Wrong working
Total saved in 12 weeks $= a_{12} = 50 + 11 \times 20 = 270$
Right working
Total saved in 12 weeks $= S_{12} = \frac{12}{2} (50 + 270) = 1920$ , that is ₹1,920
Do this: Underline 'in the nth' (use $a_{n}$ ) versus 'in n' or 'total' (use $S_{n}$ ) before you start.
Usually costs: Often 1 to 2 marks in a case-study part.
Mistake: Dropping brackets when $d$ is negative in $S_{n}$
Students write $(n - 1) d$ without brackets around the negative $d$ , and the multiplication is lost.
Wrong working
$17, 15, 13, \dots$ : $S_{10} = 5 [34 + 9 - 2] = 5 \times 41 = 205$
Right working
$S_{10} = 5 [34 + 9 (- 2)] = 5 \times 16 = 80$
Do this: Always put brackets around $d$ when you substitute, especially a negative one.
Usually costs: The answer mark, and any later part that uses this sum.

NCERT exercise guide: Arithmetic Progressions

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 5.1

What it tests: Deciding whether a list of numbers, or a real-life situation such as a fare, a saving scheme or a volume of liquid, gives an AP. You also write the first few terms from given $a$ and $d$ , read off $a$ and $d$ from a list, and find the next terms.

How to approach it: Find the differences $a_{2} - a_{1}$ , $a_{3} - a_{2}$ , $a_{4} - a_{3}$ . If they are all equal, the list is an AP and that value is $d$ . For a situation, first write the first three or four values as plain numbers, then test the differences. Simplify surds and fractions to a common form before you compare them.

Like the questions in Exercise 5.1 that ask whether a list is an AP and what its next terms are

Check whether the list $3, 12, 27, 48, \dots$ is an AP. If it is, write its common difference and its next two terms.

Show the solutionSolution

Simplify each surd: $12 = 23$ , $27 = 33$ , $48 = 43$ .
So the list is $3, 23, 33, 43, \dots$
Differences: $23 - 3 = 3$ , $33 - 23 = 3$ , $43 - 33 = 3$ .
All differences are equal, so it is an AP with $a = 3$ and $d = 3$ .
Next terms: $43 + 3 = 53 = 75$ and $53 + 3 = 63 = 108$ .

Answer: Yes, it is an AP with common difference $3$ . The next two terms are $75$ and $108$ (that is, $53$ and $63$ ).

Exercise 5.2

What it tests: Using $a_{n} = a + (n - 1) d$ : finding a given term, filling in missing terms, finding which term equals a number, testing whether a number is a term, finding the AP from two conditions, counting terms and the nth term from the end, and simple word problems.

How to approach it: Write every given condition as an equation in $a$ and $d$ . Subtracting two such equations usually removes $a$ and gives $d$ at once, because $a_{m} - a_{k} = (m - k) d$ . When you solve for $n$ , the answer must be a natural number; if it is not, the number is not a term. The formula itself comes from noticing that the nth term is the first term plus $(n - 1)$ steps of size $d$ .

Like the questions in Exercise 5.2 that give two terms and ask for another term

The 9th term of an AP is 47 and its 15th term is 77. Find the 30th term. Which term of this AP is 247?

Show the solutionSolution

$a + 8 d = 47$ ... (1) and $a + 14 d = 77$ ... (2).
Subtract (1) from (2): $6 d = 30$ , so $d = 5$ .
From (1): $a = 47 - 40 = 7$ .
$a_{30} = 7 + 29 \times 5 = 7 + 145 = 152$ .
For 247: $7 + (n - 1) \times 5 = 247$ gives $(n - 1) \times 5 = 240$ , so $n - 1 = 48$ and $n = 49$ .
Check: $a_{15} = 7 + 14 \times 5 = 77$ , which matches the data.

Answer: The 30th term is 152, and 247 is the 49th term.

Exercise 5.3

What it tests: Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ and $S_{n} = \frac{n}{2} (a + l)$ : finding sums, finding a missing one of $a$ , $d$ , $n$ , $l$ , $S_{n}$ from the others, finding $n$ from a given sum, finding the nth term from an expression for $S_{n}$ , and daily-life word problems such as prizes, savings, rows of plants and stacks.

How to approach it: List what is given among $a$ , $d$ , $n$ , $a_{n}$ and $S_{n}$ , then pick the formula that uses them. Use $S_{n} = \frac{n}{2} (a + l)$ when the last term is known. In word problems, decide clearly what $a$ , $d$ and $n$ stand for before writing any equation, and reject roots that are negative or not whole numbers.

Like the prize and reward word problems in Exercise 5.3

A school wants to give a total of ₹3,600 as 8 cash prizes to students for their attendance. Each prize is ₹50 less than the prize just before it. Find the value of each prize.

Show the solutionSolution

The prizes form an AP with $n = 8$ , $d = - 50$ and $S_{8} = 3600$ . Let the first prize be $a$ .
$S_{8} = \frac{8}{2} [2 a + 7 (- 50)] = 4 (2 a - 350)$ .
So $4 (2 a - 350) = 3600$ , giving $2 a - 350 = 900$ .
$2 a = 1250$ , so $a = 625$ .
The prizes are 625, 575, 525, 475, 425, 375, 325, 275.
Check: $S_{8} = \frac{8}{2} (625 + 275) = 4 \times 900 = 3600$ .

Answer: The prizes are ₹625, ₹575, ₹525, ₹475, ₹425, ₹375, ₹325 and ₹275.

Arithmetic Progressions: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5 marks per paper; 55 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5 marks per paper; 44 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Arithmetic Progressions

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q37 · 4 marks · Case-basedCompetency-based

Case Study – 2

In a garden, saplings of rose flowers were planted at equal intervals to form a spiral pattern. The spiral is made up of successive semicircles, with centres alternatively at A and B, starting with centre at A, of radii 50 cm, 100 cm, 150 cm, ....... as shown in the figure given below. Spiral 1 has 10 flowers, Spiral 2 has 20 flowers, Spiral 3 has 30 flowers and so on.

Based on the above information, answer the following questions :

(i)
What is the radius of the $1 3^{t h}$ spiral ?
1 mark
(ii)
If the radius of the $n^{t h}$ spiral is 500 cm, find the value of n.
1 mark
(iii)
(a) Find the total number of saplings till the $1 1^{t h}$ spiral.
2 marks
OR
(iii)
(b) Till which spiral, will there be a total of 450 saplings ?
2 marks

Figure in the paper: Spiral made of successive semicircles on a horizontal line with points A and B on it, centres alternating. Arcs labelled l1 (small, above line, around A-B), l2 (below line), l3 (above, larger), l4 (below, larger), and further arcs continuing outward as a dotted curve on the right. See it in the official paper.

Try it first, then show our solutionSolution

(i) The radii 50, 100, 150, … form an AP with $a = 50$ and $d = 50$ , so $r_{n} = 50 n$ . Hence $r_{13} = 50 + 12 \times 50 = 650$ cm.
1 mark
(ii) $50 + (n - 1) 50 = 500 \Rightarrow 50 n = 500 \Rightarrow n = 10$ .
1 mark
(iii)(a) The numbers of flowers 10, 20, 30, … form an AP with $a = 10$ and $d = 10$ . Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ : $S_{11} = \frac{11}{2} [2 (10) + 10 (10)]$ .
1 mark
(iii)(a) $S_{11} = \frac{11}{2} \times 120 = 660$ saplings.
1 mark
(iii)(b) $\frac{n}{2} [20 + (n - 1) 10] = 450 \Rightarrow 5 n (n + 1) = 450 \Rightarrow n^{2} + n - 90 = 0$ .
1 mark
(iii)(b) $(n + 10) (n - 9) = 0 \Rightarrow n = 9$ , since $n = - 10$ is rejected because $n$ must be positive. So there are 450 saplings till the 9th spiral.
1 mark

Answer: (i) 650 cm; (ii) n = 10; (iii)(a) 660 saplings; (iii)(b) 9th spiral

Where the marks are

(i) Uses the AP of radii to get $r_{13} = 650$ cm (1 mark)
(ii) Solves $50 n = 500$ to get $n = 10$ (1 mark)
(iii)(a) Correct substitution in $S_{11}$ (1 mark)
(iii)(a) Total = 660 saplings (1 mark)
(iii)(b) Forms $n^{2} + n - 90 = 0$ (1 mark)
(iii)(b) Solves to get $n = 9$ , rejecting the negative root (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/4/1, Q7 · 1 mark · MCQ

The sum of first $n$ terms of an A.P. is $502$ . If the first and the last terms are $2$ and $192$ respectively, then the value of $n$ is :

(A) $10$
(B) $5$
(C) $15$
(D) $20$

Try it first, then show our solutionSolution

Use $S_{n} = \frac{n}{2} (a + l)$ .
$502 = \frac{n}{2} (2 + 192) = 10 n 2 \Rightarrow n = 5$ , so option (B).
1 mark

Answer: n = 5, option (B)

Where the marks are

Uses $S_{n} = \frac{n}{2} (a + l)$ to get $n = 5$ (1 mark)

See it in the full 2026 board paper, set 430/4/1

2026 Basic, set 430/4/1, Q32 · 5 marks · Long answer

The third and ninth term of an A.P. are $4$ and $-8$ respectively.

(i)
Which term of the A.P. is zero ?
0 marks
(ii)
Find the value of $n$ if $S_{n} = -36$ .
0 marks

Try it first, then show our solutionSolution

(i) $a_{3} = a + 2 d = 4$ ...(1)
½ mark
$a_{9} = a + 8 d = - 8$ ...(2)
½ mark
Subtracting (1) from (2): $6 d = - 12 \Rightarrow d = - 2$ ; then $a = 4 - 2 (- 2) = 8$ .
1 mark
$a_{n} = 0 \Rightarrow 8 + (n - 1) (- 2) = 0 \Rightarrow n - 1 = 4 \Rightarrow n = 5$ . So the 5th term is zero.
1 mark
(ii) $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = - 36 \Rightarrow \frac{n}{2} [16 + (n - 1) (- 2)] = - 36$
1 mark
$\Rightarrow \frac{n}{2} (18 - 2 n) = - 36 \Rightarrow 9 n - n^{2} = - 36 \Rightarrow n^{2} - 9 n - 36 = 0$
½ mark
$\Rightarrow (n - 12) (n + 3) = 0 \Rightarrow n = 12$ or $n = - 3$ . Since n is a natural number, reject $n = - 3$ ; hence $n = 12$ .
½ mark

Answer: (i) 5th term; (ii) n = 12

Where the marks are

Writes $a + 2 d = 4$ (½ mark)
Writes $a + 8 d = - 8$ (½ mark)
Solves to get $d = - 2$ , $a = 8$ (½ + ½ marks)
Gets $n = 5$ from $a_{n} = 0$ (1 mark)
Correct $S_{n}$ equation (1 mark)
Forms $n^{2} - 9 n - 36 = 0$ (½ mark)
Factorises and takes $n = 12$ , rejecting $- 3$ (½ mark)

See it in the full 2026 board paper, set 430/4/1

Every Arithmetic Progressions question, by year

Standard (66 questions, sample papers included)

For 2027

Sample paper 2026-27: Q5 (1 mark, solved), Q28(A) (3 marks, solved), Q28(B) (3 marks, solved)

2026

Set 30/1/1: Q6 (1 mark, solved), Q36 (4 marks, solved)

Set 30/2/1: Q7 (1 mark, solved), Q20 (1 mark, solved), Q36 (4 marks, solved)

Set 30/3/1: Q6 (1 mark, solved), Q11 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved)

Set 30/4/1: Q3 (1 mark, solved), Q37 (4 marks, solved)

Set 30/5/1: Q8 (1 mark, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Sample paper 2025-26: Q21(A) (2 marks, solved), Q21(B) (2 marks, solved), Q36 (4 marks, solved)

2025

Set 30/1/1: Q11 (1 mark, solved), Q36 (4 marks, solved)

Set 30/2/1: Q15 (1 mark, solved), Q19 (1 mark, solved), Q31 (3 marks, solved)

Set 30/3/1: Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 30/4/1: Q37 (4 marks, solved)

Set 30/5/1: Q20 (1 mark, solved), Q37 (4 marks, solved)

Set 30/6/1: Q37 (4 marks, solved)

Sample paper 2024-25: Q4 (1 mark, solved), Q36 (4 marks, solved)

2024

Set 30/1/1: Q4 (1 mark, solved), Q33(A) (5 marks, solved), Q33(B) (5 marks, solved)

Set 30/2/1: Q3 (1 mark, solved), Q16 (1 mark, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 30/3/1: Q2 (1 mark, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 30/4/1: Q12 (1 mark, solved), Q37 (4 marks, solved)

Set 30/5/1: Q1 (1 mark, solved), Q11 (1 mark, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Sample paper 2023-24: Q5 (1 mark, solved), Q20 (1 mark, solved), Q36 (4 marks, solved)

2023

Set 30/1/1: Q3 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 30/2/1: Q32 (5 marks, solved)

Set 30/4/1: Q3 (1 mark, solved), Q20 (1 mark, solved), Q26(A) (3 marks, solved), Q26(B) (3 marks, solved)

Set 30/5/1: Q3 (1 mark, solved), Q14 (1 mark, solved), Q27 (3 marks, solved)

Set 30/6/1: Q18 (1 mark, solved), Q34(A) (5 marks, solved), Q34(B) (5 marks, solved)

Basic (55 questions, sample papers included)

For 2027

Sample paper 2026-27: Q6 (1 mark, solved), Q7 (1 mark, solved), Q37 (4 marks, solved)

2026

Set 430/1/1: Q37 (4 marks, solved)

Set 430/2/1: Q37 (4 marks, solved)

Set 430/3/1: Q36 (4 marks, solved)

Set 430/4/1: Q7 (1 mark, solved), Q14 (1 mark, solved), Q32 (5 marks, solved)

Set 430/5/1: Q10 (1 mark, solved), Q14 (1 mark, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Sample paper 2025-26: Q5 (1 mark, solved), Q6 (1 mark, solved), Q31 (3 marks, solved)

2025

Set 430/1/1: Q37 (4 marks, solved)

Set 430/2/1: Q37 (4 marks, solved)

Set 430/3/1: Q36 (4 marks, solved)

Set 430/4/1: Q3 (1 mark, solved), Q10 (1 mark, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Set 430/5/1: Q6 (1 mark, solved), Q38 (4 marks, solved)

Set 430/6/1: Q2 (1 mark, solved), Q16 (1 mark, solved), Q31(a) (3 marks, solved)

Sample paper 2024-25: Q10 (1 mark, solved), Q19 (1 mark, solved), Q23(A) (2 marks, solved), Q23(B) (2 marks, solved)

2024

Set 430/1/1: Q2 (1 mark, solved), Q33 (5 marks, solved)

Set 430/2/1: Q4 (1 mark, solved), Q36 (4 marks, solved)

Set 430/3/1: Q17 (1 mark, solved), Q36 (4 marks, solved)

Set 430/4/1: Q10 (1 mark, solved), Q35 (5 marks, solved)

Set 430/5/1: Q20 (1 mark, solved), Q37 (4 marks, solved)

Sample paper 2023-24: Q36 (4 marks, solved)

2023

Set 430/1/1: Q6 (1 mark, solved), Q36 (4 marks, solved)

Set 430/2/1: Q1 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 430/4/1: Q33(A) (5 marks, solved), Q33(B) (5 marks, solved)

Set 430/5/1: Q1 (1 mark, solved), Q12 (1 mark, solved), Q25 (2 marks, solved)

Set 430/6/1: Q16 (1 mark, solved), Q33 (5 marks, solved)

Practice questions on Arithmetic Progressions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQHarder board level
The sum of the first $n$ terms of an AP is $3 n^{2} + 5 n$ . If its $k$ th term is 164, then $k$ is
(a) 26
(b) 27
(c) 28
(d) 82
Hint
Find $a_{n} = S_{n} - S_{n - 1}$ first, then set it equal to 164.
Answer
(b) 27. $a_{n} = (3 n^{2} + 5 n) - [3 (n - 1)^{2} + 5 (n - 1)] = 6 n + 2$ . Check: $a_{1} = 8 = S_{1}$ . Then $6 k + 2 = 164$ gives $k = 27$ .
Q2 · 2 marks · Very short answerBoard level
If $2 p + 1$ , $13$ and $5 p - 3$ are three consecutive terms of an AP, find $p$ and write the three terms.
Hint
The middle term is the average of the other two, so $2 b = a + c$ .
Answer
$2 \times 13 = (2 p + 1) + (5 p - 3)$ gives $26 = 7 p - 2$ , so $p = 4$ . The terms are 9, 13, 17 (common difference 4).
Q3 · 2 marks · Very short answerBoard level
Find the sum $50 + 46 + 42 + \dots + 10$ .
Hint
Find the number of terms first, then use $S_{n} = \frac{n}{2} (a + l)$ .
Answer
$a = 50$ , $d = - 4$ , $l = 10$ : $n = \frac{10 - 50}{- 4} + 1 = 11$ . $S_{11} = \frac{11}{2} (50 + 10) = 330$ .
Q4 · 3 marks · Short answerBoard level
The 5th term of an AP is 19, and its 13th term is 20 more than its 8th term. Find the AP and its 20th term.
Hint
$a_{13} - a_{8} = 5 d$ , so the second condition gives $d$ straight away.
Answer
$5 d = 20$ , so $d = 4$ . $a + 16 = 19$ , so $a = 3$ . The AP is $3, 7, 11, 15, \dots$ and $a_{20} = 3 + 19 \times 4 = 79$ .
Q5 · 5 marks · Long answerHarder board level
Starting from the definition of an AP, derive the formula $a_{n} = a + (n - 1) d$ . Hence prove that if the $p$ th term of an AP is $q$ and the $q$ th term is $p$ (where $p \neq = q$ ), then its $(p + q)$ th term is zero.
Hint
Subtract the two term equations to get $d$ , then find $a$ in terms of $p$ and $q$ .
Answer
$a_{2} = a + d$ , $a_{3} = a + 2 d$ , and each step adds one $d$ , so the $n$ th term has $(n - 1)$ steps: $a_{n} = a + (n - 1) d$ . From $a + (p - 1) d = q$ and $a + (q - 1) d = p$ : $(p - q) d = q - p$ , so $d = - 1$ and $a = p + q - 1$ . Then $a_{p + q} = (p + q - 1) + (p + q - 1) (- 1) = 0$ .
Q6 · 5 marks · Long answerHarder board level
The sum of the first 9 terms of an AP is 162, and the ratio of its 6th term to its 13th term is $1 : 2$ . Find the first term, the common difference and the sum of the first 15 terms. Which term of the AP is three times its 4th term?
Hint
Cross-multiply the ratio to get a link between $a$ and $d$ , then use $S_{9}$ .
Answer
$2 (a + 5 d) = a + 12 d$ gives $a = 2 d$ . $S_{9} = 9 (a + 4 d) = 162$ gives $a + 4 d = 18$ , so $6 d = 18$ , $d = 3$ , $a = 6$ . $S_{15} = \frac{15}{2} (12 + 42) = 405$ . $a_{4} = 15$ , and $6 + 3 (n - 1) = 45$ gives $n = 14$ , so it is the 14th term.
Q7 · 3 marks · Short answerBoard level
How many terms of the AP $27, 24, 21, \dots$ must be taken so that their sum is zero? Explain, using the terms themselves, why the sum comes out to zero.
Hint
Set $S_{n} = 0$ and remember that $n \neq = 0$ .
Answer
$\frac{n}{2} [54 - 3 (n - 1)] = 0$ gives $57 - 3 n = 0$ , so $n = 19$ . The 10th term is $27 - 27 = 0$ and the 19th term is $- 27$ , so the terms pair off as $27$ with $- 27$ , $24$ with $- 24$ , and so on, around 0, and cancel.
Q8 · 5 marks · Long answerHarder board level
Aman repays a loan of ₹1,18,000 in monthly instalments. The first instalment is ₹1,000, and each instalment is ₹100 more than the one before. How many instalments does he pay, and what is the last instalment?
Hint
The total paid is $S_{n}$ with $a = 1000$ and $d = 100$ . Divide the equation by 50 before factorising.
Answer
$\frac{n}{2} [2000 + 100 (n - 1)] = 118000$ gives $n^{2} + 19 n - 2360 = 0$ , that is $(n + 59) (n - 40) = 0$ , so $n = 40$ . Last instalment $= 1000 + 39 \times 100 =$ ₹4,900. Check: $20 \times (1000 + 4900) = 118000$ .
Q9 · 3 marks · Short answerHOTS / competency
Priya says that if you keep adding the terms of the AP $5, 9, 13, 17, \dots$ , the running total will at some point be exactly 1000. Is she right? Justify your answer.
Hint
Write $S_{n}$ in terms of $n$ and ask whether $S_{n} = 1000$ has a natural-number solution.
Answer
No. $S_{n} = \frac{n}{2} [10 + 4 (n - 1)] = n (2 n + 3)$ . Setting this equal to 1000 gives $2 n^{2} + 3 n - 1000 = 0$ , whose discriminant is 8009, not a perfect square, so $n$ is not a natural number. Also, $S_{21} = 945$ and $S_{22} = 1034$ , so the total jumps past 1000.
Q10 · 4 marks · Case-basedHOTS / competency
A school quiz has 10 rounds. The points available in the rounds form an AP, the last round carries 42 points, and the total for all 10 rounds is 240 points. (i) Find the points for the first round and the increase from one round to the next. (ii) A team scored full points in each of the first $k$ rounds and nothing after that, for a total of 96. Find $k$ .
Hint
Use $S_{10} = \frac{10}{2} (a + l)$ first, then set $S_{k} = 96$ .
Answer
(i) $5 (a + 42) = 240$ , so $a = 6$ , and $6 + 9 d = 42$ gives $d = 4$ . (ii) $S_{k} = \frac{k}{2} [12 + 4 (k - 1)] = 2 k^{2} + 4 k = 96$ , so $k^{2} + 2 k - 48 = 0$ , that is $(k + 8) (k - 6) = 0$ , and $k = 6$ . Check: $3 (6 + 26) = 96$ .
Q11 · 3 marks · Short answerHarder board level
The 3rd term of an AP is 10 and its 7th term is 22. Find the sum of its terms from the 11th term to the 30th term.
Hint
Find $a$ and $d$ first. The block of terms is $S_{30} - S_{10}$ , or a new AP of 20 terms starting at $a_{11}$ .
Answer
$a + 2 d = 10$ and $a + 6 d = 22$ give $4 d = 12$ , so $d = 3$ and $a = 4$ . $a_{11} = 4 + 30 = 34$ and $a_{30} = 4 + 87 = 91$ , so the block has 20 terms with sum $\frac{20}{2} (34 + 91) = 1250$ . Check: $S_{30} - S_{10} = 15 (8 + 87) - 5 (8 + 27) = 1425 - 175 = 1250$ .
Q12 · 3 marks · Short answerHarder board level
In an AP, the sum of the first $m$ terms is equal to the sum of the first $n$ terms, where $m \neq = n$ . Prove that the sum of its first $(m + n)$ terms is zero.
Hint
Write $S_{m} = S_{n}$ , bring everything to one side and take out the factor $(m - n)$ .
Answer
$\frac{m}{2} [2 a + (m - 1) d] = \frac{n}{2} [2 a + (n - 1) d]$ gives $2 a (m - n) + d [(m^{2} - n^{2}) - (m - n)] = 0$ , that is $(m - n) [2 a + (m + n - 1) d] = 0$ . Since $m \neq = n$ , $2 a + (m + n - 1) d = 0$ . So $S_{m + n} = \frac{m + n}{2} [2 a + (m + n - 1) d] = 0$ .
Q13 · 3 marks · Short answerHOTS / competency
For a list of numbers, the sum of the first $n$ terms is $S_{n} = n^{2} + 2 n + 3$ . Rahul says: " $a_{n} = S_{n} - S_{n - 1} = 2 n + 1$ , so the list is an AP with common difference 2." Find his error, write the first three terms, and decide whether the list is an AP.
Hint
The rule $a_{n} = S_{n} - S_{n - 1}$ works only for $n \geq 2$ . Find $a_{1}$ from $S_{1}$ .
Answer
$a_{n} = 2 n + 1$ holds only for $n \geq 2$ ; the first term is $a_{1} = S_{1} = 6$ , not 3. Since $S_{2} = 11$ and $S_{3} = 18$ , the terms are $6, 5, 7, \dots$ with differences $- 1$ and $2$ , so the list is not an AP. The constant term 3 in $S_{n}$ is the warning sign: the $S_{n}$ of an AP has no constant term.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the common difference of an AP?Show answer
The fixed number $d = a_{k + 1} - a_{k}$ , the same for every pair of consecutive terms. It can be positive, negative or zero.
2.State the formula for the nth term.Show answer
$a_{n} = a + (n - 1) d$
3.When are $a$ , $b$ , $c$ consecutive terms of an AP?Show answer
When $2 b = a + c$ .
4.Give both forms of the sum of the first n terms.Show answer
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ and $S_{n} = \frac{n}{2} (a + l)$
5.How do you get the nth term from the sum?Show answer
$a_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$ , and $a_{1} = S_{1}$ .
6.What is the nth term from the end of an AP with last term l?Show answer
$l - (n - 1) d$
7.Is $5, 5, 5, 5, \dots$ an AP?Show answer
Yes. It is an AP with $d = 0$ .
8.If $a_{n} = 7 - 3 n$ , what is d?Show answer
$d = - 3$ , the coefficient of $n$ .

Best order to study it

Learn what an AP is, how to find $d$ (later minus earlier), and the condition $2 b = a + c$ . Practise finding $k$ .
Derive $a_{n} = a + (n - 1) d$ yourself. Practise finding terms, testing whether a number is a term, counting terms, terms from the end and middle terms.
Practise forming two equations in $a$ and $d$ from given conditions, including the $a - d$ , $a$ , $a + d$ choice for three terms.
Derive the sum formula by writing the sum forwards and backwards. Then practise sums of $n$ terms and sums of multiples in a range.
Practise finding $n$ from a given sum (quadratics, rejecting roots) and finding $a_{n}$ from a given $S_{n}$ .
Finish with word problems and case studies, deciding each time whether the question needs $a_{n}$ or $S_{n}$ .

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you derive $a_{n} = a + (n - 1) d$ and explain why it is $n - 1$ , not $n$ ?
Can you derive $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ by adding the sum to itself written backwards?
Can you find $k$ when three expressions in $k$ are in AP?
Can you decide whether a given number is a term of an AP and justify your answer?
Can you count the multiples of 6 between 100 and 500 without missing the $+ 1$ ?
Can you find the 12th term from the end and the middle term of a finite AP?
Can you find $a$ and $d$ when two terms, or a term and a sum, are given?
Can you find $n$ when the sum is given, and explain a double answer when $d$ is negative?
Can you find $a_{n}$ and $d$ from an $S_{n}$ such as $3 n^{2} + 4 n$ , and check with $a_{1} = S_{1}$ ?
In a word problem, can you tell at a glance whether the question wants $a_{n}$ or $S_{n}$ ?

Arithmetic Progressions: frequently asked questions

Do I need to learn the derivations of the nth term and sum formulas?

Yes. The 2026-27 syllabus lists the derivation of the nth term and of the sum of the first n terms. Practise writing each derivation in a few clear lines: the term pattern for the nth term, and the forwards-plus-backwards addition for the sum.

Which sum formula should I use, the one with d or the one with the last term?

Use the one with the last term when the first and last terms are both known, as in sums of multiples, because it is quicker. Use the one with d when you know the first term, the common difference and the number of terms, or when you must form a quadratic in n.

Why do I sometimes get two values of n for the same sum?

This happens when the common difference is negative. After some point the terms become negative, and a group of terms can add up to zero. So the total returns to the same value. If both values are natural numbers, both are valid, and you should say why in one line.

Can the common difference be a fraction or negative?

Yes. The common difference can be any real number: positive for an increasing AP, negative for a decreasing one, zero for a constant one, or a fraction. Only the number of terms n has to be a natural number.

How can I check my answer quickly in the exam?

Substitute back. Write the first few terms and check the given conditions, or put n equal to 1 in your nth term and compare with the first term or with S1. For sums, recompute using the other sum formula.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)