MathsCBSE Class 10 · Arithmetic Progressions

Introduction to Arithmetic Progressions for Class 10

Updated 14 October 2026 · Fact-checked

An arithmetic progression is a list of numbers in which you add the same fixed number, called the common difference, to each term to get the next term. The board asks you to find the common difference, test whether a list is an AP, find an unknown so three terms are in AP, and spot APs in real situations.

In an AP every step is the same d (here +3); check every consecutive pair, because 1, 3, 5, 8 breaks at the last step.

Illustration drawn for QuestionBuddy.

Understand Introduction to Arithmetic Progressions

A list of numbers written in a definite order is called a sequence. Each number is a term: $a_{1}$ is the first term, $a_{2}$ the second, and so on. An arithmetic progression (AP) is a list in which you get every term after the first by adding the same fixed number to the term before it. That fixed number is the common difference $d$ . An AP with a last term, such as the 12 monthly instalments of a loan, is a finite AP; one that goes on forever, such as 3, 7, 11, ..., is an infinite AP.

So $d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots$ . The difference can be positive ( $2, 5, 8, \dots$ with $d = 3$ ), negative ( $10, 6, 2, \dots$ with $d = - 4$ ), zero ( $7, 7, 7, \dots$ ), or a fraction ( $1, \frac{3}{2}, 2, \dots$ with $d = \frac{1}{2}$ ). Always subtract later minus earlier, or you get the sign of $d$ wrong.

Why does this matter? Once you know the first term $a$ and $d$ , the whole AP is fixed: $a, a + d, a + 2 d, a + 3 d, \dots$ . This general form is the starting point for the $n$ th term formula and the sum formula you meet next in the chapter. Every later question begins by finding $a$ and $d$ .

To test whether a list is an AP, find the difference of every pair of consecutive terms. Equal differences everywhere means an AP. One unequal difference means it is not. Checking just the first pair is not enough, because a list like $1, 3, 5, 8$ starts like an AP and then breaks. If the terms are given by a rule $a_{n}$ , compute $a_{n + 1} - a_{n}$ : the list is an AP exactly when this does not depend on $n$ .

For three consecutive terms $a$ , $b$ , $c$ of an AP, $b - a = c - b$ , so $2 b = a + c$ . The middle term is the average of its neighbours. This is the fastest way to find an unknown $k$ when the terms contain it. It also lets you insert numbers between two given numbers.

APs appear in real life whenever something changes by a fixed amount each step: a fare rising by the same amount per km, a fixed monthly increase in savings, seats rising by a fixed number per row. If the change is a fixed multiple (doubling, 10% growth), it is not an AP.

Introduction to Arithmetic Progressions: formulas to remember

Common difference: $d = a_{2} - a_{1} = a_{3} - a_{2} = a_{k + 1} - a_{k}$; Later term minus earlier term. Must be the same for every consecutive pair.
General form of an AP: $a, a + d, a + 2 d, a + 3 d, \dots$; An AP is fixed completely by $a$ and $d$ .
Three terms in AP: $2 b = a + c$; For consecutive terms $a$ , $b$ , $c$ . Use it to find an unknown $k$ .
Test from a rule: $a_{n + 1} - a_{n} = constant (independent of n)$; If $a_{n}$ is given as an expression, the list is an AP only if this difference has no $n$ in it.

How to solve Introduction to Arithmetic Progressions questions

Use this method for any question that asks you to find $d$ , test a list, or find an unknown so that terms are in AP.

1Write down the terms clearly, simplifying any surds or fractions first (for example $8 = 22$ ).
2Find each difference as later term minus earlier term.
3If all differences are equal, state that the list is an AP with that $d$ ; if any differs, state that it is not an AP and show the two unequal differences.
4If the terms are given by a rule $a_{n}$ , compute $a_{n + 1} - a_{n}$ and check whether $n$ cancels.
5If an unknown $k$ appears in three consecutive terms, use $2 \times middle = first + third$ and solve for $k$ .
6Substitute back, write the terms, and check that both differences are equal.
7Write the final answer as a sentence, giving $a$ , $d$ and the next terms if asked.

Introduction to Arithmetic Progressions 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.

Three terms in AP (2b = a + c)

Imagine you take three unit tests in maths and want your marks to rise by the same amount each time. You scored 54 in Test 1 and plan to score 78 in Test 3.

What must you score in Test 2 so the three marks are in AP? If the marks keep rising by the same amount, will Test 5 exceed 100? If instead the marks are $3 x + 6$ , $5 x + 4$ and $8 x$ , find $x$ and the three marks.

Try it first, then show the solutionSolution

Let Test 2 be $b$ . Then $2 b = 54 + 78 = 132$ , so $b = 66$ .
Check: $66 - 54 = 12$ and $78 - 66 = 12$ .
Continuing with $d = 12$ : Test 4 would be 90 and Test 5 would be 102, which exceeds the maximum of 100, so the pattern cannot continue to Test 5.
For the second part: $2 (5 x + 4) = (3 x + 6) + 8 x$ , so $10 x + 8 = 11 x + 6$ and $x = 2$ .
Marks: $3 (2) + 6 = 12$ , $5 (2) + 4 = 14$ , $8 (2) = 16$ . Check: $14 - 12 = 2 = 16 - 14$ .

Answer: You need 66 marks in Test 2; continuing the AP, Test 5 would need 102, more than 100, so the pattern must stop. In the second case $x = 2$ and the marks are 12, 14 and 16.

Memory hook: The middle test is the average of the other two.

Tricks to remember Introduction to Arithmetic Progressions

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

Later minus earlier
Think of d as 'how much you step forward'. Step from the earlier term to the later one: $d =$ later $-$ earlier.
Use it when: Stops sign errors in decreasing APs like $20, 17, 14$ where $d = - 3$ .
Middle is the average
In three AP terms, the middle sits exactly halfway: $b = \frac{a + c}{2}$ , so $2 b = a + c$ .
Use it when: Finding $k$ in one line.
Add, not multiply
AP = Add the same. Doubling or percentage growth multiplies, so it is not an AP.
Use it when: Spotting APs in word problems and assertion-reason items.
Check all gaps
One bad gap breaks the AP. Check every consecutive pair given.
Use it when: Testing lists in MCQs.

Introduction to Arithmetic Progressions: solved examples

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

Example 1: Testing lists for an AP · 1 mark · MCQBoard level

Which of the following is an AP?
(a) $2, 8, 18, 32$
(b) $1, 4, 9, 16$
(c) $2, 4, 8, 16$
(d) $3, 6, 9, 12$

Try it first, then show the solutionSolution

(a) Simplify: $2, 22, 32, 42$ ; differences all $2$ , so AP.
(b) Differences 3, 5, 7 are unequal, so not an AP.
(c) Differences 2, 4, 8 are unequal, so not an AP.
(d) $6 - 3 \neq = 9 - 6$ (about 0.72 and 0.55), so not an AP.

Answer: (a) $2, 8, 18, 32$ , an AP with $d = 2$ .

Example 2: Finding k with squared terms · 3 marks · Short answerBoard level

Find the value of $k$ for which $k^{2} + 4 k + 8$ , $2 k^{2} + 3 k + 6$ and $3 k^{2} + 4 k + 4$ are three consecutive terms of an AP. Also write the terms.

Try it first, then show the solutionSolution

Three terms in AP satisfy $2 b = a + c$ . (½)
$2 (2 k^{2} + 3 k + 6) = (k^{2} + 4 k + 8) + (3 k^{2} + 4 k + 4)$ (½)
$4 k^{2} + 6 k + 12 = 4 k^{2} + 8 k + 12$ (½)
$6 k = 8 k$ , so $k = 0$ . (½)
Terms: 8, 6, 4. (½)
Check: $6 - 8 = - 2$ and $4 - 6 = - 2$ , equal. (½)

Answer: $k = 0$ , and the terms are 8, 6, 4 with common difference $- 2$ .

Example 3: AP from a rule · 3 marks · Short answerHarder board level

The $n$ th term of a list is $a_{n} = 5 - 3 n$ . Show that the list is an AP and find its first term and common difference. Is the list with $b_{n} = n^{2} + 1$ an AP? Justify.

Try it first, then show the solutionSolution

$a_{n + 1} = 5 - 3 (n + 1) = 2 - 3 n$ . (½)
$a_{n + 1} - a_{n} = (2 - 3 n) - (5 - 3 n) = - 3$ , independent of $n$ , so it is an AP. (1)
First term $a_{1} = 5 - 3 = 2$ , and $d = - 3$ . (½)
$b_{n + 1} - b_{n} = (n + 1)^{2} + 1 - n^{2} - 1 = 2 n + 1$ , which depends on $n$ . (½)
So $b_{n}$ is not an AP (terms 2, 5, 10, ... have differences 3, 5). (½)

Answer: $a_{n} = 5 - 3 n$ is an AP with $a = 2$ , $d = - 3$ (terms 2, $- 1$ , $- 4$ , ...). $b_{n} = n^{2} + 1$ is not an AP since its difference $2 n + 1$ changes.

Example 4: Three numbers in AP with sum and product · 3 marks · Short answerHarder board level

Three numbers are in AP. Their sum is 24 and their product is 440. Find the numbers.

Try it first, then show the solutionSolution

Let the numbers be $a - d$ , $a$ , $a + d$ . (½)
Sum: $3 a = 24$ , so $a = 8$ . (½)
Product: $(8 - d) (8) (8 + d) = 440$ . (½)
$64 - d^{2} = 55$ , so $d^{2} = 9$ , $d = \pm 3$ . (1)
Numbers: 5, 8, 11 (or 11, 8, 5). Check: sum 24, product $5 \times 8 \times 11 = 440$ . (½)

Answer: The numbers are 5, 8, 11 (or in reverse order 11, 8, 5).

Example 5: Case study: two plans for festival diyas · 4 marks · Case-basedHOTS / competency

Suppose your school decorates four steps with diyas for a festival. Plan A puts $4 n + 8$ diyas on step $n$ . Plan B puts 10, 13, 18 and 25 diyas on steps 1 to 4.
(i) Write the Plan A numbers for steps 1 to 4 and decide which of the two plans forms an AP. (1)
(ii) Which plan uses more diyas in total on the four steps, and by how many? (1)
(iii) A third plan, Plan C, forms an AP with 16 diyas on step 2 and 28 diyas on step 4. Find its first term and common difference, list its four terms, and compare its total with Plan A. (2)

Try it first, then show the solutionSolution

(i) Plan A: $4 (1) + 8 = 12$ , then 16, 20, 24; each difference is 4, so Plan A is an AP. (½)
Plan B: differences $13 - 10 = 3$ , $18 - 13 = 5$ , $25 - 18 = 7$ are unequal, so Plan B is not an AP. (½)
(ii) Plan A total $= 12 + 16 + 20 + 24 = 72$ ; Plan B total $= 10 + 13 + 18 + 25 = 66$ . Plan A uses 6 more. (1)
(iii) Let the first term be $a$ and the common difference $d$ : $a + d = 16$ and $a + 3 d = 28$ . (½)
Subtracting, $2 d = 12$ , so $d = 6$ and $a = 10$ . (½)
Plan C: 10, 16, 22, 28; check: every difference is 6. (½)
Total $= 10 + 16 + 22 + 28 = 76$ , which is $76 - 72 = 4$ more than Plan A. (½)

Answer: (i) Plan A: 12, 16, 20, 24, an AP with $d = 4$ ; Plan B is not an AP. (ii) Plan A, by 6 diyas (72 against 66). (iii) $a = 10$ , $d = 6$ ; Plan C is 10, 16, 22, 28 and uses 76 diyas, 4 more than Plan A.

Example 6: Writing an AP from a and d · 3 marks · Short answerBoard level

Write the first four terms of the AP in each case: (i) $a = - 2$ , $d = 0$ (ii) $a = 4$ , $d = - 3$ (iii) $a = \frac{1}{2}$ , $d = - \frac{1}{6}$ . Say whether each AP is increasing, decreasing or constant.

Try it first, then show the solutionSolution

Use the general form $a, a + d, a + 2 d, a + 3 d$ . (½)
(i) Adding 0 each time gives $- 2, - 2, - 2, - 2$ : a constant AP. (½)
(ii) $4, 4 - 3, 4 - 6, 4 - 9$ (½)
gives $4, 1, - 2, - 5$ : a decreasing AP. (½)
(iii) In sixths: $\frac{3}{6}, \frac{3}{6} - \frac{1}{6}, \frac{3}{6} - \frac{2}{6}, \frac{3}{6} - \frac{3}{6}$ (½)
gives $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}, 0$ : a decreasing AP. (½)

Answer: (i) $- 2, - 2, - 2, - 2$ (constant). (ii) $4, 1, - 2, - 5$ (decreasing). (iii) $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}, 0$ (decreasing).

Example 7: Two conditions, then a membership check · 3 marks · Short answerHarder board level

In an AP, the sum of the 3rd and 5th terms is 4, and the 2nd term is 6 more than the 4th term. Find the first term and the common difference, write the first five terms, and decide whether $- 61$ and $61$ are terms of this AP.

Try it first, then show the solutionSolution

$a_{2} - a_{4} = (a + d) - (a + 3 d) = - 2 d = 6$ , so $d = - 3$ . (½)
$a_{3} + a_{5} = 2 a + 6 d = 4$ , so $2 a - 18 = 4$ and $a = 11$ . (½)
First five terms: 11, 8, 5, 2, $- 1$ . (½)
$- 61$ : $11 - 3 (n - 1) = - 61$ gives $n - 1 = 24$ , $n = 25$ , a natural number, so $- 61$ is the 25th term. (1)
$61$ : $11 - 3 (n - 1) = 61$ gives $n - 1 = - \frac{50}{3}$ , not possible, so 61 is not a term (the AP is decreasing from 11). (½)

Answer: $a = 11$ , $d = - 3$ ; AP 11, 8, 5, 2, $- 1$ , ...; $- 61$ is the 25th term; 61 is not a term.

Example 8: Reciprocal terms: proving no value works · 3 marks · Short answerHOTS / competency

Show that there is no real number $x$ for which $\frac{1}{x + 1}$ , $\frac{1}{x + 2}$ and $\frac{1}{x + 3}$ are three consecutive terms of an AP.

Try it first, then show the solutionSolution

If they were in AP, $2 \times \frac{1}{x + 2} = \frac{1}{x + 1} + \frac{1}{x + 3}$ , with $x \neq = - 1, - 2, - 3$ . (½)
Right side: $\frac{( x + 3 ) + ( x + 1 )}{( x + 1 ) ( x + 3 )} = \frac{2 ( x + 2 )}{( x + 1 ) ( x + 3 )}$ . (½)
So $\frac{2}{x + 2} = \frac{2 ( x + 2 )}{( x + 1 ) ( x + 3 )}$ ; cross-multiplying, $(x + 1) (x + 3) = (x + 2)^{2}$ . (1)
$x^{2} + 4 x + 3 = x^{2} + 4 x + 4$ , so $3 = 4$ , which is false. (½)
No value of $x$ satisfies the condition, so the three terms can never be in AP. (½)

Answer: No real $x$ works, because the AP condition reduces to $3 = 4$ , which is impossible.

Example 9: AP from a linear rule with two unknowns · 5 marks · Long answerHarder board level

The $n$ th term of a list is $a_{n} = p n + q$ , where $p$ and $q$ are constants.
(i) Show that the list is an AP and find its common difference.
(ii) If the 3rd term is 11 and the 7th term is 27, find $p$ and $q$ and write the first four terms.
(iii) Is 75 a term of the list? Is 150? Justify.

Try it first, then show the solutionSolution

(i) $a_{n + 1} - a_{n} = [p (n + 1) + q] - [p n + q] = p$ , which does not depend on $n$ , so it is an AP with $d = p$ . (1)
(ii) $a_{3} = 3 p + q = 11$ and $a_{7} = 7 p + q = 27$ . (½)
Subtracting, $4 p = 16$ , so $p = 4$ ; then $q = 11 - 12 = - 1$ . (1)
So $a_{n} = 4 n - 1$ , and the first four terms are 3, 7, 11, 15. (½)
(iii) $4 n - 1 = 75$ gives $4 n = 76$ , $n = 19$ , a natural number, so 75 is the 19th term. (1)
$4 n - 1 = 150$ gives $n = \frac{151}{4}$ , not a natural number, so 150 is not a term. (1)

Answer: (i) AP with $d = p$ . (ii) $p = 4$ , $q = - 1$ ; terms 3, 7, 11, 15. (iii) 75 is the 19th term; 150 is not a term.

Where students lose marks in Introduction to Arithmetic Progressions

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

Mistake: Subtracting in the wrong order
Students subtract earlier minus later, flipping the sign of d.
Wrong working
For $20, 17, 14$ : $d = 20 - 17 = 3$ .
Right working
$d = 17 - 20 = - 3$ .
Do this: Always compute later term minus earlier term.
Usually costs: The full mark in an MCQ and every later step that uses d.
Mistake: Checking only one pair
The first two terms may differ by a constant that does not continue.
Wrong working
$1, 3, 5, 8$ : $3 - 1 = 2$ , so it is an AP with $d = 2$ .
Right working
$3 - 1 = 2$ , $5 - 3 = 2$ , $8 - 5 = 3$ ; unequal, so not an AP.
Do this: Check every consecutive pair given and show them.
Usually costs: The answer mark in testing questions.
Mistake: Not simplifying surds first
Unsimplified surds hide equal differences.
Wrong working
$3, 12, 27$ cannot be compared, so not an AP.
Right working
$3, 23, 33$ : AP with $d = 3$ .
Do this: Write each surd in simplest form before subtracting.
Usually costs: 1 mark in Section A.
Mistake: Using 2a = b + c with the wrong term in the middle
Students double the first term instead of the middle one.
Wrong working
For $2 p + 1, 13, 5 p - 3$ : $2 (2 p + 1) = 13 + 5 p - 3$ , so $p = - 8$ .
Right working
$2 (13) = (2 p + 1) + (5 p - 3)$ , so $p = 4$ .
Do this: Double the middle term only, then substitute back to check.
Usually costs: Usually all 2 marks.

Exam tips

Show at least two differences when you claim a list is an AP; one difference alone may lose the reasoning mark.
When the terms contain an unknown, use 2 × middle = first + third and always substitute back to show the terms.
For three numbers in AP with a given sum, take a − d, a, a + d so that d cancels; this saves time in 3-mark questions.
In assertion-reason items, test the claim with actual differences; a rule like n squared never gives an AP.
Fractions and surds often appear in MCQs: simplify each term before subtracting.

Practice questions on Introduction to 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 · MCQBoard level
The common difference of the AP $\frac{1}{p}, \frac{1 - p}{p}, \frac{1 - 2 p}{p}, \dots$ is
(a) $p$
(b) $- p$
(c) $- 1$
(d) $1$
Hint
Subtract the first term from the second and simplify.
Answer
(c) $- 1$ , since $\frac{1 - p}{p} - \frac{1}{p} = \frac{- p}{p} = - 1$ .
Q2 · 2 marks · Very short answerBoard level
For what value of $p$ are $p - 1$ , $2 p + 1$ , $5 p - 3$ three consecutive terms of an AP? Find the common difference and the 5th term of this AP.
Hint
Use $2 b = a + c$ , then keep adding $d$ .
Answer
$2 (2 p + 1) = (p - 1) + (5 p - 3)$ , so $4 p + 2 = 6 p - 4$ and $p = 3$ . Terms 2, 7, 12 with $d = 5$ ; continuing, 17, 22, so the 5th term is 22.
Q3 · 3 marks · Short answerBoard level
Show that $(a - b)^{2}$ , $a^{2} + b^{2}$ , $(a + b)^{2}$ are in AP and find the common difference.
Hint
Expand the squares and find both differences.
Answer
Terms: $a^{2} - 2 ab + b^{2}$ , $a^{2} + b^{2}$ , $a^{2} + 2 ab + b^{2}$ . Both differences equal $2 ab$ , so they are in AP with $d = 2 ab$ .
Q4 · 3 marks · Short answerHarder board level
Three numbers in AP have sum 15 and the sum of their squares is 83. Find the numbers.
Hint
Take $a - d$ , $a$ , $a + d$ ; the sum gives $a$ .
Answer
$3 a = 15$ , $a = 5$ . $(5 - d)^{2} + 25 + (5 + d)^{2} = 83$ gives $75 + 2 d^{2} = 83$ , $d = \pm 2$ . Numbers 3, 5, 7 (or 7, 5, 3).
Q5 · 1 mark · Assertion-reasonBoard level
Assertion (A): The list given by $a_{n} = 2 n + 3$ is an AP with common difference 2.
Reason (R): If $a_{n} = p n + q$ for constants $p$ and $q$ , the list is an AP with common difference $p$ .
(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.
Hint
Compute $a_{n + 1} - a_{n}$ for the general rule $p n + q$ .
Answer
(a). For $p n + q$ , $a_{n + 1} - a_{n} = p$ , so R is true. Here $p = 2$ , giving terms 5, 7, 9, ... with $d = 2$ , so A is true and follows directly from R.
Q6 · 1 mark · MCQBoard level
Which of the following rules gives an AP?
(a) $a_{n} = n^{2} - n$
(b) $a_{n} = 2^{n}$
(c) $a_{n} = 7 - 4 n$
(d) $a_{n} = \frac{1}{n}$
Hint
Write the first three or four terms of each and compare differences, or find $a_{n + 1} - a_{n}$ .
Answer
(c) $a_{n} = 7 - 4 n$ , since $a_{n + 1} - a_{n} = - 4$ for every $n$ (terms 3, $- 1$ , $- 5$ , ...). The others are not APs: (a) terms 0, 2, 6; (b) terms 2, 4, 8; (d) terms 1, $\frac{1}{2}$ , $\frac{1}{3}$ do not have a constant difference.
Q7 · 2 marks · Very short answerBoard level
If $k + 2$ , $4 k - 6$ and $3 k - 2$ are three consecutive terms of an AP, find $k$ and the three terms.
Hint
Double the middle term and equate to the sum of the other two.
Answer
$2 (4 k - 6) = (k + 2) + (3 k - 2)$ , so $8 k - 12 = 4 k$ and $k = 3$ . Terms 5, 6, 7 with $d = 1$ .
Q8 · 4 marks · Case-basedHOTS / competency
Suppose a student saves money for a school trip. She saves ₹50 in week 1, ₹65 in week 2, ₹80 in week 3, and continues in the same pattern.
(i) Show that her weekly savings form an AP and state the common difference. (1)
(ii) How much does she save in week 6? (1)
(iii) Her friend's weekly savings also form an AP, with ₹70 in week 3 and ₹110 in week 7. Find the friend's week 1 saving and common difference, and find the week in which the student saves exactly ₹30 more than the friend. (2)
Hint
Use differences for (i); for (iii), write two equations in $a$ and $d$ , then track how the gap grows each week.
Answer
(i) $65 - 50 = 15$ and $80 - 65 = 15$ are equal, so it is an AP with $d = 15$ (₹15 more each week). (1) (ii) Weeks 1 to 6: 50, 65, 80, 95, 110, 125, so ₹125. (1) (iii) Friend: $a + 2 d = 70$ and $a + 6 d = 110$ give $4 d = 40$ , $d = 10$ , $a = 50$ , so ₹50 in week 1 and ₹10 more each week. (1) Both save ₹50 in week 1, and the gap grows by ₹5 a week, so a gap of ₹30 needs $30 \div 5 = 6$ more weeks: week 7. Check: student ₹140, friend ₹110. (1)
Q9 · 3 marks · Short answerHarder board level
In an AP, the sum of the 2nd and 5th terms is 23, and the 4th term is 1 less than twice the 2nd term. Find the first term, the common difference and the first five terms.
Hint
Write each term as $a + (steps) \times d$ and form two equations.
Answer
$(a + d) + (a + 4 d) = 23$ gives $2 a + 5 d = 23$ . $a + 3 d = 2 (a + d) - 1$ gives $d = a - 1$ . Then $2 a + 5 (a - 1) = 23$ , so $7 a = 28$ , $a = 4$ , $d = 3$ . AP: 4, 7, 10, 13, 16. Check: $7 + 16 = 23$ and $13 = 2 (7) - 1$ .
Q10 · 3 marks · Short answerHarder board level
Find all values of $x$ for which $2 x - 1$ , $x^{2}$ and $3 x + 4$ are three consecutive terms of an AP. For each value, write the three terms and check. Which value gives terms that are all whole numbers?
Hint
Use $2 b = a + c$ ; you get a quadratic.
Answer
$2 x^{2} = (2 x - 1) + (3 x + 4) = 5 x + 3$ , so $2 x^{2} - 5 x - 3 = 0$ , $(2 x + 1) (x - 3) = 0$ , $x = 3$ or $x = - \frac{1}{2}$ . For $x = 3$ : 5, 9, 13 ( $d = 4$ ). For $x = - \frac{1}{2}$ : $- 2$ , $\frac{1}{4}$ , $\frac{5}{2}$ ( $d = \frac{9}{4}$ ). Both give APs; only $x = 3$ gives whole numbers.
Q11 · 4 marks · Case-basedHOTS / competency
Case study: A student makes a row of squares with matchsticks. One square uses 4 sticks, two squares in a row use 7, three use 10, and so on.
(i) Show that the numbers of sticks form an AP and write its first term and common difference. (1 mark)
(ii) How many sticks are needed for a row of 20 squares? (1 mark)
(iii) With 100 sticks, what is the longest row of squares she can make, and are any sticks left over? She then makes square grids instead: a 1 × 1 grid uses 4 sticks, a 2 × 2 grid 12 and a 3 × 3 grid 24. Is this second list an AP? Justify. (2 marks)
Hint
Each new square in a row needs 3 more sticks. For the grids, look at the differences.
Answer
(i) 4, 7, 10, ...: differences are all 3, so it is an AP with $a = 4$ , $d = 3$ . (1)
(ii) $a_{20} = 4 + 19 \times 3 = 61$ sticks. (1)
(iii) $4 + 3 (n - 1) \leq 100$ gives $n - 1 \leq 32$ , so $n = 33$ squares, using exactly $4 + 96 = 100$ sticks, none left over. Grids: 4, 12, 24 have differences 8 and 12, which are not equal, so this list is not an AP. (2)

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 an AP?Show answer
A list where each term after the first is got by adding the same fixed number d.
2.How do you find d?Show answer
Later term minus earlier term, for example a₂ − a₁.
3.Can d be negative or zero?Show answer
Yes. Negative gives a decreasing AP; zero gives a constant list.
4.Condition for a, b, c in AP?Show answer
2b = a + c.
5.General form of an AP?Show answer
a, a + d, a + 2d, a + 3d, ...
6.Is 3, 9, 27 an AP?Show answer
No; differences 6 and 18 are unequal.

Introduction to Arithmetic Progressions: frequently asked questions

Is a constant list like 5, 5, 5 an AP?

Yes. Its common difference is zero, and each term is got by adding zero to the one before. It is a valid AP.

How many differences should I check?

Check every consecutive pair that is given. If the list is defined by a rule, show that the difference between the next term and the current term does not depend on n.

What is the difference between an AP and a list that doubles?

An AP adds the same amount each time. A list that doubles multiplies by the same number, so its differences keep changing and it is not an AP.

Is Exercise 5.4 part of the exam?

Exercise 5.4 is marked optional in the NCERT book and is not assessed. Practise Exercises 5.1 to 5.3 for the board exam.