NCERT Exemplar Solutions Class 10 Maths Arithmetic Progressions — Exercise 5.2

NCERT Exemplar Problems, Class 10 Mathematics — Arithmetic Progressions, Exercise 5.2. All 8 questions solved.

This exercise asks one question eight times over: is this list an AP, and why? The verdict is worth little; the justification is the answer. And the justification is always the same calculation — compute $a_2 – a_1$, $a_3 – a_2$, $a_4 – a_3$ and see whether they agree.

Two points decide almost every part:

• One equal pair of differences is not enough. A list is an AP only if every consecutive difference is the same. Question 2 exists entirely to make this point, and the trap is a real one: the first two differences agree and the third does not.
• A sequence given by a formula is an AP exactly when that formula is linear in $n$. If $a_n = pn + q$ then $a_{n+1} – a_n = p$, a constant. Anything with $n^2$ in it is not an AP. Question 8 is this test.

Key insight. “Justify your answer” means show the differences. Writing “yes, it is an AP” earns nothing; writing $a_2 – a_1 = a_3 – a_2 = a_4 – a_3 = 11$ earns everything.

Question 1

Which of the following form an AP? Justify your answer.

(i) $-1$, $-1$, $-1$, $-1$, …

AP. Every difference is $-1 – (-1) = 0$. A constant list is an AP with $d = 0$ — the definition asks for a fixed difference, and zero is a perfectly good fixed number.

(ii) $0$, $2$, $0$, $2$, …

Not an AP. The differences are $2$, $-2$, $2$, … which alternate rather than stay fixed.

(iii) $1$, $1$, $2$, $2$, $3$, $3$, …

Not an AP. The differences are $0$, $1$, $0$, $1$, $0$ — again not constant. Each value repeating twice is enough to break it.

(iv) $11$, $22$, $33$, …

AP, with $d = 11$: $22 – 11 = 33 – 22 = 11$. These are the multiples of $11$, and the multiples of any fixed number always form an AP.

(v) $\dfrac12$, $\dfrac13$, $\dfrac14$, …

Not an AP.

$$a_2 – a_1 = \frac13 – \frac12 = -\frac16, \qquad a_3 – a_2 = \frac14 – \frac13 = -\frac{1}{12}$$

The gaps shrink as the denominators grow. It is worth noticing that the denominators $2, 3, 4$ do form an AP — but the question is about the terms themselves.

(vi) $2$, $2^2$, $2^3$, $2^4$, …

Not an AP. The list is $2, 4, 8, 16$, whose differences are $2$, $4$, $8$ — doubling, not constant. Each term is twice the one before, which makes this a geometric progression.

(vii) $\sqrt3$, $\sqrt{12}$, $\sqrt{27}$, $\sqrt{48}$, …

AP. Simplify each surd first — that is the whole question:

$$\sqrt{12} = 2\sqrt3, \qquad \sqrt{27} = 3\sqrt3, \qquad \sqrt{48} = 4\sqrt3$$

so the list is $\sqrt3$, $2\sqrt3$, $3\sqrt3$, $4\sqrt3$ with a constant difference $\sqrt3$. Left unsimplified, $3$, $12$, $27$, $48$ look like a quadratic pattern and the AP is invisible.

Question 2

Justify whether it is true to say that $-1$, $-\dfrac32$, $-2$, $\dfrac52$, … forms an AP as $a_2 – a_1 = a_3 – a_2$.

Solution. Not true — and the reason given in the question is exactly the flaw in the reasoning.

$$a_2 – a_1 = -\frac32 – (-1) = -\frac12$$ $$a_3 – a_2 = -2 – \left(-\frac32\right) = -\frac12$$ $$a_4 – a_3 = \frac52 – (-2) = \frac92$$

The first two differences do agree, which is what makes the claim tempting. But $a_4 – a_3 = \frac92 \ne -\frac12$, so the list is not an AP.

The general point: an AP needs $a_{k+1} – a_k$ to be the same for every $k$, not for one value of $k$. Checking two differences establishes nothing about the third.

Question 3

For the AP $-3$, $-7$, $-11$, …, can we find $a_{30} – a_{20}$ directly without actually finding $a_{30}$ and $a_{20}$? Give reasons for your answer.

Solution. Yes. Subtracting two terms of an AP cancels the first term:

$$a_{30} – a_{20} = \left[a + 29d\right] – \left[a + 19d\right] = 10d$$

Here $d = -7 – (-3) = -4$, so

$$a_{30} – a_{20} = 10 \times (-4) = -40$$

The general rule is $a_p – a_q = (p – q)d$, and it holds for any AP whatever its first term — which is precisely why the individual terms are not needed.

Question 4

Two APs have the same common difference. The first term of one AP is $2$ and that of the other is $7$. The difference between their $10$th terms is the same as the difference between their $21$st terms, which is the same as the difference between any two corresponding terms. Why?

Solution. Let the shared common difference be $d$. The $n$-th terms of the two APs are

$$2 + (n-1)d \qquad\text{and}\qquad 7 + (n-1)d$$

Subtracting, the $(n-1)d$ cancels for every $n$:

$$\left[7 + (n-1)d\right] – \left[2 + (n-1)d\right] = 7 – 2 = 5$$

So the difference is $5$ whatever $n$ is — at the 10th terms, at the 21st terms, and at every other position.

Geometrically the two APs advance in lockstep: they take the same step each time, so the gap set by their starting points never changes.

Question 5

Is $0$ a term of the AP $31$, $28$, $25$, …? Justify your answer.

Solution. No. Here $a = 31$ and $d = -3$. If $0$ were the $n$-th term then

$$0 = 31 + (n-1)(-3) \implies 3(n-1) = 31 \implies n – 1 = \frac{31}{3} \implies n = \frac{34}{3}$$

A term number must be a positive integer, and $\frac{34}{3}$ is not one. So $0$ never appears.

Concretely, the AP runs $31, 28, \dots, 4, 1, -2, \dots$ — it steps straight over zero, from $1$ to $-2$.

Question 6

The taxi fare after each km, when the fare is ₹15 for the first km and ₹8 for each additional km, does not form an AP as the total fare (in ₹) after each km is $15$, $8$, $8$, $8$, … Is the statement true? Give reasons.

Solution. Not true, and the error is in the list rather than in the conclusion drawn from it. The question asks for the total fare after each kilometre, which accumulates:

$$\text{after 1 km: } 15, \quad \text{after 2 km: } 15 + 8 = 23, \quad 31, \quad 39, \ \dots$$

The list $15$, $8$, $8$, $8$ is the extra charged for each successive kilometre, not the total.

The correct list has differences $23 – 15 = 8$, $31 – 23 = 8$, $39 – 31 = 8$ — constant. So the total fare does form an AP, with $a = 15$ and $d = 8$.

Question 7

In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.

(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is ₹400.

AP. The list is $400$, $400$, $400$, … with $d = 0$. Constant, and therefore an AP.

(ii) The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is ₹250 and it increases by ₹50 for the next higher class.

AP. The list is $250$, $300$, $350$, …, increasing by a fixed ₹50 each time, so $d = 50$.

(iii) The amount of money in the account of Varun at the end of every year when ₹1000 is deposited at simple interest of $10\%$ per annum.

AP. Simple interest is calculated on the original principal every year, so the same ₹100 is added annually:

$$1100, \quad 1200, \quad 1300, \quad 1400, \ \dots$$

with $d = 100$. This is the one to contrast with compound interest, where the interest itself earns interest and the amounts $1100$, $1210$, $1331$ have differences $110$, $121$ — not an AP.

(iv) The number of bacteria in a certain food item after each second, when they double every second.

Not an AP. Doubling gives $b$, $2b$, $4b$, $8b$, … whose differences are $b$, $2b$, $4b$ — themselves doubling. This is a geometric progression.

Question 8

Justify whether it is true to say that the following are the $n$th terms of an AP.

A list is an AP exactly when $a_{n+1} – a_n$ is free of $n$ — which happens precisely when $a_n$ is a linear expression in $n$.

(i) $2n – 3$

True. The terms are $-1$, $1$, $3$, $5$, … and

$$a_{n+1} – a_n = \left[2(n+1) – 3\right] – \left[2n – 3\right] = 2$$

a constant. So it is an AP with $a = -1$ and $d = 2$ — and note that $d$ is the coefficient of $n$.

(ii) $3n^2 + 5$

Not true. The terms are $8$, $17$, $32$, $53$, … with differences $9$, $15$, $21$ — increasing. In general

$$a_{n+1} – a_n = 3\left[(n+1)^2 – n^2\right] = 3(2n+1) = 6n + 3$$

which depends on $n$, so it is not constant.

(iii) $1 + n + n^2$

Not true. The terms are $3$, $7$, $13$, $21$, … with differences $4$, $6$, $8$. Generally

$$a_{n+1} – a_n = \left[(n+1) + (n+1)^2\right] – \left[n + n^2\right] = 2n + 2$$

again depending on $n$.

Both (ii) and (iii) are quadratic in $n$, and a quadratic $n$-th term always gives differences that themselves form an AP — a second-order pattern, not a first-order one.

Common mistakes

• Checking only one pair of differences. Question 2 is built on this. Compute at least three differences, and say in the answer that you did.
• Leaving surds unsimplified in Question 1(vii). $\sqrt3$, $\sqrt{12}$, $\sqrt{27}$, $\sqrt{48}$ looks like nothing until you write it as $\sqrt3$, $2\sqrt3$, $3\sqrt3$, $4\sqrt3$.
• Reading “total fare” as “fare for that kilometre” in Question 6. The statement’s own list is what is wrong; the conclusion happens to be wrong too, and saying only “the statement is false” without correcting the list misses the point.
• Confusing simple with compound interest in Question 7(iii). Simple interest adds a fixed amount each year and gives an AP; compound interest multiplies and does not.
• Treating $d = 0$ as disqualifying. Questions 1(i) and 7(i) are both constant lists, and both are APs.
• Concluding “not an AP” from a formula without testing it. $2n – 3$ contains no obvious “$+d$”, but it is an AP. The test is whether the expression is linear in $n$; if it is, the coefficient of $n$ is the common difference.
• Looking at the denominators in Question 1(v). $2$, $3$, $4$ is an AP, but $\frac12$, $\frac13$, $\frac14$ is not. The reciprocals of an AP are almost never an AP.

Practise next

• Exemplar Exercise 5.3 Q1–18 — verifying APs and finding their terms, which is this exercise made computational.
• Exemplar Exercise 5.1 for the objective versions of Questions 3 and 4.
• NCERT textbook Exercise 5.1 — the book’s own “does this situation form an AP?” questions, including taxi fares and compound interest, which Questions 6 and 7 above extend.
• NCERT textbook Exercise 5.2 for the $n$-th term, where the “is $0$ a term?” test of Question 5 is set several times.