MathsCBSE Class 9 · Sequences and Progressions

Arithmetic Progressions: nth Term and Sums, Class 9

Q: Can the common difference be negative?

Yes. Then the sequence decreases, like 25, 23, 21, \dots.

Q: Is a constant sequence an AP?

Yes, with common difference zero.

Q: Why is the formula for the sum symmetrical?

Adding the sequence to its reverse makes every pair sum to the same value a + l.

Updated 17 October 2026 · Fact-checked

An arithmetic progression (AP) has a constant common difference $d$, so the nth term is $a_n = a + (n-1)d$. The sum of the first $n$ terms is $S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}(a + l)$, where $l$ is the last term. For the natural numbers, $1 + 2 + \dots + n = \frac{n(n+1)}{2}$. Two known terms give $d$ from $a_p - a_q = (p-q)d$.

Each step adds d = 5, and first + last, second + second-last and so on all make 26, so Sum = (number of terms × pair total) ÷ 2.

Illustration drawn for QuestionBuddy.

Understand Arithmetic Progressions: nth Term and Sum of Terms

A sequence is an ordered list of numbers, called terms. In an arithmetic progression each term is obtained from the previous one by adding the same number $d$ , the common difference: $a, a + d, a + 2 d, \dots$ . Here $a = a_{1}$ is the first term and $d = a_{n + 1} - a_{n}$ may be positive, negative or zero.

The nth term. The $n$ th term is $a_{n} = a + (n - 1) d$ , because $d$ has been added $(n - 1)$ times. For the AP $7, 11, 15, \dots$ : $a = 7$ , $d = 4$ , so $a_{20} = 7 + 19 \times 4 = 83$ . To find which term equals $151$ , solve $7 + 4 (n - 1) = 151$ , which gives $n = 37$ . If the equation gives a fraction or a negative number, the value is not a term of the AP: $301$ is not in $5, 11, 17, \dots$ because $5 + 6 (n - 1) = 301$ gives $n - 1 = \frac{296}{6}$ , not a whole number.

Finding the AP from two terms. If $a_{p}$ and $a_{q}$ are known, subtract: $a_{p} - a_{q} = (p - q) d$ . With $a_{4} = 18$ and $a_{9} = 43$ : $5 d = 25$ , $d = 5$ , $a = 18 - 15 = 3$ , and the AP is $3, 8, 13, 18, \dots$ ; check: $a_{9} = 3 + 40 = 43$ . Three numbers in AP can be taken as $a - d$ , $a$ , $a + d$ : this makes their sum $3 a$ immediately.

**Sum of the first $n$ natural numbers.** $S = 1 + 2 + \dots + n$ . Write it again in reverse and add the two lists term by term: every pair sums to $n + 1$ , and there are $n$ pairs, so $2 S = n (n + 1)$ and $S = \frac{n ( n + 1 )}{2}$ . The sum of the first $50$ natural numbers is $\frac{50 \times 51}{2} = 1275$ . The same pairing trick works for any AP: the first term plus the last term equals the second plus the second-last, and so on, so
$ $S_{n} = \frac{n}{2} (a + l) = \frac{n}{2} [2 a + (n - 1) d] .$ $
The sum of the first $15$ odd numbers ( $a = 1$ , $d = 2$ ) is $\frac{15}{2} (2 + 28) = 225$ .

Counting terms in a range. The multiples of $7$ with two digits run from $14$ to $98$ . Number of terms $= \frac{98 - 14}{7} + 1 = 13$ and their sum is $\frac{13}{2} (14 + 98) = 728$ . For the multiples of $7$ between $100$ and $500$ : the first is $105$ , the last $497$ , so there are $\frac{497 - 105}{7} + 1 = 57$ terms with sum $\frac{57}{2} (105 + 497) = 17157$ .

From the sum to the terms. If $S_{n}$ is given as a formula, then $a_{1} = S_{1}$ and $a_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$ . For $S_{n} = 3 n^{2} + 2 n$ : $a_{1} = 5$ , $a_{2} = S_{2} - S_{1} = 16 - 5 = 11$ , so $d = 6$ and $a_{n} = 6 n - 1$ . This also shows that a polynomial sum of this kind belongs to an AP.

Context problems. Savings that grow by a fixed amount each week, rows of seats that increase by a fixed number, bricks stacked in decreasing rows: all are APs. Identify $a$ , $d$ , whether the question asks for a term (use $a_{n}$ ) or a total (use $S_{n}$ ), and whether $n$ must be a whole number.

Arithmetic Progressions: nth Term and Sum of Terms: formulas to remember

nth term: $a_{n} = a + (n - 1) d$; $d$ = common difference.
Common difference: $d = a_{n + 1} - a_{n}$; Constant for an AP.
Two terms: $a_{p} - a_{q} = (p - q) d$; Subtract to get $d$ quickly.
Sum of first n naturals: $1 + 2 + \dots + n = \frac{n ( n + 1 )}{2}$; Pairing first and last.
Sum of an AP: $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l)$; $l$ = last term.
Terms from the sum: $a_{n} = S_{n} - S_{n - 1}, a_{1} = S_{1}$; For $n \geq 2$ .
Number of terms in a range: $n = \frac{l - a}{d} + 1$; Must be a whole number.

How to solve Arithmetic Progressions: nth Term and Sum of Terms questions

Write $a$ , $d$ and what is asked; then pick the term or sum formula.

1Identify the first term $a$ and the common difference $d$ (subtract consecutive terms).
2For a term, use $a_{n} = a + (n - 1) d$ ; to test membership, solve for $n$ and check that it is a positive whole number.
3For an AP from two terms, subtract the equations to find $d$ , then find $a$ .
4For a sum, use $S_{n} = \frac{n}{2} (a + l)$ when the last term is known, or $\frac{n}{2} [2 a + (n - 1) d]$ otherwise.
5For three numbers in AP use $a - d, a, a + d$ when their sum is given.
6In context problems, define $n$ , round up where needed, and check the answer in the original situation.

Arithmetic Progressions: nth Term and Sum of Terms in real life

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.

Sum of an AP

Suppose a mason builds a triangular brick stack with 25 bricks in the bottom row, and each row above has 2 fewer bricks, ending with a single brick on top.

How many rows are there and how many bricks are used?

Try it first, then show the solutionSolution

Rows: $25, 23, 21, \dots, 1$ , with $d = - 2$ .
$1 = 25 + (n - 1) (- 2)$ gives $n - 1 = 12$ , so $n = 13$ .
Total $= \frac{13}{2} (25 + 1) = 169$ bricks.

Answer: There are 13 rows and 169 bricks (which is $1 3^{2}$ ).

Memory hook: Average of first and last, times the number of rows.

Tricks to remember Arithmetic Progressions: nth Term and Sum of Terms

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

Add d n minus 1 times
To reach the nth term you step $(n - 1)$ times, not $n$ times.
Use it when: Avoiding off-by-one errors in $a_{n}$ .
Average of ends times count
The sum of an AP is the average of the first and last term times the number of terms.
Use it when: Sums with a known last term.
Middle term, plus and minus d
Three numbers in AP: $a - d$ , $a$ , $a + d$ . Four: $a - 3 d$ , $a - d$ , $a + d$ , $a + 3 d$ .
Use it when: Sum and product problems.

Arithmetic Progressions: nth Term and Sum of Terms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: nth term and position · 3 marks · Short answerBoard level

For the AP $7, 11, 15, \dots$ find the 20th term, and find which term equals 151.

Try it first, then show the solutionSolution

$a = 7$ , $d = 11 - 7 = 4$ . (½)
$a_{20} = 7 + 19 \times 4 = 7 + 76 = 83$ . (1)
$7 + 4 (n - 1) = 151$ gives $4 (n - 1) = 144$ , so $n - 1 = 36$ and $n = 37$ . (1½)

Answer: $a_{20} = 83$ and 151 is the 37th term.

Example 2: AP from two terms · 3 marks · Short answerBoard level

The 4th term of an AP is 18 and the 9th term is 43. Find the AP and its 20th term.

Try it first, then show the solutionSolution

$a + 3 d = 18$ and $a + 8 d = 43$ . (1)
Subtract: $5 d = 25$ , so $d = 5$ and $a = 18 - 15 = 3$ . (1)
The AP is $3, 8, 13, 18, \dots$ and $a_{20} = 3 + 19 \times 5 = 98$ . (1)

Answer: The AP is $3, 8, 13, 18, \dots$ ; $a_{20} = 98$ .

Example 3: Multiples of 7 · 5 marks · Long answerHarder board level

Find the sum of all two-digit multiples of 7, and the sum of all multiples of 7 between 100 and 500.

Try it first, then show the solutionSolution

Two-digit multiples of 7: $14, 21, \dots, 98$ ; $a = 14$ , $d = 7$ , $l = 98$ . (1)
$n = \frac{98 - 14}{7} + 1 = 13$ . (½)
$S = \frac{13}{2} (14 + 98) = 13 \times 56 = 728$ . (1)
Between 100 and 500: first $105$ , last $497$ (since $500 = 7 \times 71 + 3$ ). (1)
$n = \frac{497 - 105}{7} + 1 = 57$ and $S = \frac{57}{2} (105 + 497) = 57 \times 301 = 17157$ . (1½)

Answer: The sums are $728$ and $17157$ .

Example 4: Three numbers in AP · 3 marks · Short answerHarder board level

Three numbers in AP have sum 27 and product 693. Find them.

Try it first, then show the solutionSolution

Let the numbers be $a - d$ , $a$ , $a + d$ ; the sum is $3 a = 27$ , so $a = 9$ . (1)
Product: $9 (81 - d^{2}) = 693$ , so $81 - d^{2} = 77$ and $d^{2} = 4$ , $d = \pm 2$ . (1)
The numbers are $7, 9, 11$ (or in reverse order); check $7 \times 9 \times 11 = 693$ . (1)

Answer: The numbers are 7, 9, 11.

Example 5: Weekly savings · 4 marks · Case-basedHOTS / competency

Case: Asha saves 50 rupees in week 1 and 15 rupees more each week than the week before. (i) How much does she save in week 12? (ii) How much has she saved in total after 12 weeks? (iii) In which week does the weekly saving first exceed 300 rupees? (iv) After how many weeks does her total first reach at least 5000 rupees?

Try it first, then show the solutionSolution

(i) $a_{12} = 50 + 11 \times 15 = 215$ rupees. (½)
(ii) $S_{12} = \frac{12}{2} (50 + 215) = 6 \times 265 = 1590$ rupees. (1)
(iii) $50 + 15 (n - 1) > 300$ gives $n - 1 > 16.67$ , so $n = 18$ (week 17 gives $290$ , week 18 gives $305$ ). (1)
(iv) $S_{n} = \frac{n}{2} [100 + 15 (n - 1)] = \frac{n ( 85 + 15 n )}{2}$ . $S_{23} = 4945 < 5000$ and $S_{24} = 5340 \geq 5000$ , so after 24 weeks. (1½)

Answer: (i) 215; (ii) 1590; (iii) week 18; (iv) 24 weeks.

Where students lose marks in Arithmetic Progressions: nth Term and Sum of Terms

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

Mistake: Using n instead of n - 1
Counting the first term as a step.
Wrong working
$a_{20} = 7 + 20 \times 4 = 87$ .
Right working
$a_{20} = 7 + 19 \times 4 = 83$ .
Do this: The first term needs no step.
Usually costs: The whole answer.
Mistake: Counting terms without adding 1
Treating the difference of ends as the count.
Wrong working
Number of two-digit multiples of 7 $= \frac{98 - 14}{7} = 12$ .
Right working
$\frac{98 - 14}{7} + 1 = 13$ .
Do this: Both ends are included: add 1.
Usually costs: 2 marks.
Mistake: Accepting a fractional n
Not checking that the position is a whole number.
Wrong working
301 is the $50.3$ rd term.
Right working
$n$ must be a whole number, so 301 is not a term.
Do this: Check $n \in N$ .
Usually costs: 1 mark.

Exam tips

Write $a$ , $d$ , $n$ before substituting.
Check the last term of a range is in the AP before counting.
Verify a derived AP by recomputing one of the given terms.
For total-vs-term questions, underline "sum" or "term" in the question.

Practice questions on Arithmetic Progressions: nth Term and Sum of Terms

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 · MCQBoard level
The sum of the first 50 natural numbers is
(a) 1250
(b) 1275
(c) 2550
(d) 1300
Hint
Use $\frac{n ( n + 1 )}{2}$ .
Answer
(b) 1275. $\frac{50 \times 51}{2} = 1275$ .
Q2 · 3 marks · Short answerBoard level
Is 301 a term of the AP $5, 11, 17, 23, \dots$ ? Is 305? Give reasons and, if a term, state its position.
Hint
Solve $5 + 6 (n - 1) =$ the number and check that $n$ is a whole number.
Answer
$5 + 6 (n - 1) = 301$ gives $n - 1 = \frac{296}{6}$ , not a whole number, so 301 is not a term. $5 + 6 (n - 1) = 305$ gives $n - 1 = 50$ , so 305 is the 51st term.
Q3 · 3 marks · Short answerHarder board level
Find $n$ if $1 + 2 + 3 + \dots + n = 210$ , and find the sum of the first 15 odd natural numbers.
Hint
$n (n + 1) = 420$ ; for the odd numbers $a = 1$ , $d = 2$ .
Answer
$\frac{n ( n + 1 )}{2} = 210$ gives $n (n + 1) = 420 = 20 \times 21$ , so $n = 20$ . Sum of 15 odd numbers $= \frac{15}{2} [2 + 14 \times 2] = \frac{15}{2} \times 30 = 225$ .
Q4 · 5 marks · Long answerHarder board level
The sum of the first $n$ terms of a sequence is $S_{n} = 3 n^{2} + 2 n$ . Find the first term, the common difference and the nth term, and show that the sequence is an AP. Find the 10th term.
Hint
$a_{1} = S_{1}$ and $a_{n} = S_{n} - S_{n - 1}$ .
Answer
$a_{1} = S_{1} = 5$ . $S_{2} = 16$ , so $a_{2} = 11$ . $a_{n} = S_{n} - S_{n - 1} = 3 n^{2} + 2 n - [3 (n - 1)^{2} + 2 (n - 1)] = 6 n - 1$ for $n \geq 2$ , and $6 (1) - 1 = 5 = a_{1}$ so the formula holds for $n = 1$ too. Differences: $a_{n + 1} - a_{n} = 6$ , constant, so it is an AP with $a = 5$ and $d = 6$ . $a_{10} = 59$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: a hall has 20 rows of seats. The first row has 18 seats and each row has 2 seats more than the one before. (i) How many seats are in the 20th row? (ii) How many seats are there in total? (iii) How many rows would be needed for at least 600 seats (with the same pattern)? (iv) Which row has exactly 40 seats?
Hint
$a = 18$ , $d = 2$ ; for (iii) compare $S_{17}$ and $S_{18}$ .
Answer
(i) $18 + 19 \times 2 = 56$ . (ii) $\frac{20}{2} (18 + 56) = 740$ . (iii) $S_{17} = \frac{17}{2} (36 + 32) = 578 < 600$ and $S_{18} = \frac{18}{2} (36 + 34) = 630 \geq 600$ , so 18 rows. (iv) $18 + 2 (n - 1) = 40$ gives $n = 12$ .

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.nth term of an AP?Show answer
$a + (n - 1) d$ .
2.Sum of the first n natural numbers?Show answer
$\frac{n ( n + 1 )}{2}$ .
3.Sum of an AP with last term l?Show answer
$\frac{n}{2} (a + l)$ .
4.How to find d from two terms?Show answer
$d = \frac{a _{p} - a _{q}}{p - q}$ .
5.Three numbers in AP with sum 27?Show answer
Take $9 - d$ , $9$ , $9 + d$ .

Arithmetic Progressions: nth Term and Sum of Terms: frequently asked questions

Can the common difference be negative?

Yes. Then the sequence decreases, like 25, 23, 21, \dots.

Is a constant sequence an AP?

Yes, with common difference zero.

Why is the formula for the sum symmetrical?

Adding the sequence to its reverse makes every pair sum to the same value a + l.