NCERT Exemplar Solutions Class 10 Maths Arithmetic Progressions — Exercise 5.1

NCERT Exemplar Problems, Class 10 Mathematics — Arithmetic Progressions, Exercise 5.1. All 18 multiple-choice questions solved.

Every question in this exercise comes out of three formulas. For an AP with first term $a$ and common difference $d$:

$$a_n = a + (n-1)d, \qquad S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}\left(a + a_n\right)$$

The second form of $S_n$ is the one to reach for whenever the last term is known — Questions 17 and 18 both fall out of it in a line.

Two shortcuts are worth learning here because the exemplar uses them repeatedly:

• Differences of terms. $a_p – a_q = (p – q)d$. The first terms cancel, so nothing else about the AP matters. Questions 9, 10 and 11 are this and nothing else.
• Counting backwards from the end. The $k$-th term from the end of an AP with last term $l$ is $l – (k-1)d$. Question 13 is one line this way and several lines the other.

Key insight. Before substituting, ask which of $a$, $d$, $n$, $a_n$, $S_n$ the question actually gives you and which it wants. Nearly every question here gives three of them, and the formula that contains exactly those three and the unknown is the one to use.

Choose the correct answer from the given four options:

Question 1

In an AP, if $d = -4$, $n = 7$, $a_n = 4$, then $a$ is

Solution. Use $a_n = a + (n-1)d$ and solve for $a$:

$$4 = a + 6(-4) = a – 24 \implies a = 28$$

Since $d$ is negative the terms decrease, so the first term must be larger than the seventh — which rules out (A) and (B) before any arithmetic.

Question 2

In an AP, if $a = 3.5$, $d = 0$, $n = 101$, then $a_n$ will be

Solution. With $d = 0$ every term equals the first:

$$a_{101} = 3.5 + 100 \times 0 = 3.5$$

A constant sequence is a perfectly good AP — the definition requires the difference between consecutive terms to be the same, not to be non-zero.

Question 3

The list of numbers $-10$, $-6$, $-2$, $2$, … is

Solution. Subtract each term from the next, in that order:

$$-6 – (-10) = 4, \qquad -2 – (-6) = 4, \qquad 2 – (-2) = 4$$

The difference is constant at $4$, so it is an AP with $d = 4$.

The numbers are increasing, so $d$ must be positive — which discards (A) and (C) at once. The sign errors those options represent come from subtracting in the wrong order.

Question 4

The $11$th term of the AP $-5$, $-\dfrac52$, $0$, $\dfrac52$, … is

Solution. Here $a = -5$ and $d = -\frac52 – (-5) = \frac52$.

$$a_{11} = -5 + 10 \times \frac52 = -5 + 25 = 20$$

Question 5

The first four terms of an AP, whose first term is $-2$ and the common difference is $-2$, are

Solution. Add $d = -2$ each time:

$$-2, \quad -2 + (-2) = -4, \quad -6, \quad -8$$

Option (D) is the geometric pattern of doubling, and (B) alternates sign; neither has a constant difference.

Question 6

The $21$st term of the AP whose first two terms are $-3$ and $4$ is

Solution. The common difference is $d = 4 – (-3) = 7$, so

$$a_{21} = -3 + 20 \times 7 = -3 + 140 = 137$$

Option (C) is $143 = -3 + 21 \times 7$ — what you get by using $n$ instead of $n – 1$, which is the single most common error in this chapter.

Question 7

If the $2$nd term of an AP is $13$ and the $5$th term is $25$, what is its $7$th term?

Solution. The gap from the 2nd term to the 5th is three common differences:

$$a_5 – a_2 = 3d = 25 – 13 = 12 \implies d = 4$$

From the 5th term to the 7th is two more:

$$a_7 = a_5 + 2d = 25 + 8 = 33$$

Finding $a$ first ($a = 13 – 4 = 9$) also works and gives $a_7 = 9 + 24 = 33$, but stepping forward from a known term is shorter.

Question 8

Which term of the AP $21$, $42$, $63$, $84$, … is $210$?

Solution. Every term is a multiple of $21$ — the AP is $21 \times 1, 21 \times 2, \dots$ — so the $n$-th term is $21n$ and

$$21n = 210 \implies n = 10$$

Formally, $a = 21$, $d = 21$, and $210 = 21 + (n-1)21$ gives $n – 1 = 9$.

Question 9

If the common difference of an AP is $5$, then what is $a_{18} – a_{13}$?

Solution. Subtracting two terms cancels the first term entirely:

$$a_{18} – a_{13} = \left[a + 17d\right] – \left[a + 12d\right] = 5d = 25$$

The general rule, worth memorising, is $a_p – a_q = (p-q)d$. Note it is $18 – 13 = 5$ steps, not $6$ — you are counting gaps, not terms.

Question 10

What is the common difference of an AP in which $a_{18} – a_{14} = 32$?

Solution. By the same rule,

$$a_{18} – a_{14} = (18 – 14)d = 4d = 32 \implies d = 8$$

Question 11

Two APs have the same common difference. The first term of one of these is $-1$ and that of the other is $-8$. Then the difference between their $4$th terms is

Solution. Write both fourth terms and subtract:

$$\left(-1 + 3d\right) – \left(-8 + 3d\right) = -1 + 8 = 7$$

The $3d$ cancels because the common difference is shared. The point of the question is that the gap between two such APs never changes — the difference between their $n$-th terms is the difference between their first terms, for every $n$.

Question 12

If $7$ times the $7$th term of an AP is equal to $11$ times its $11$th term, then its $18$th term will be

Solution. Write out the condition:

$$7(a + 6d) = 11(a + 10d)$$ $$7a + 42d = 11a + 110d$$ $$-4a = 68d \implies a = -17d$$

Now the 18th term:

$$a_{18} = a + 17d = -17d + 17d = 0$$

This is a standard and worth-remembering result: if $p\,a_p = q\,a_q$ with $p \ne q$, then $a_{p+q} = 0$. Here $7 + 11 = 18$.

Question 13

The $4$th term from the end of the AP $-11$, $-8$, $-5$, …, $49$ is

Solution. Here $d = 3$ and the last term is $l = 49$. Counting backwards from the end, the $k$-th term from the end is $l – (k-1)d$:

$$\text{4th from the end} = 49 – 3 \times 3 = 49 – 9 = 40$$

Option (C) is $49 – 2 \times 3$, the error of counting the last term as the first step rather than as step zero. There is no need to find how many terms the AP has.

Question 14

The famous mathematician associated with finding the sum of the first $100$ natural numbers is

Solution. Carl Friedrich Gauss, as a schoolboy, is said to have summed $1$ to $100$ almost instantly by pairing the numbers from the two ends: $1 + 100$, $2 + 99$, $3 + 98$, and so on — fifty pairs each totalling $101$, giving $50 \times 101 = 5050$.

That pairing argument is exactly the derivation of $S_n = \frac{n}{2}(a + a_n)$, so the anecdote is the formula in disguise.

Question 15

If the first term of an AP is $-5$ and the common difference is $2$, then the sum of the first $6$ terms is

Solution.

$$S_6 = \frac{6}{2}\left[2(-5) + 5(2)\right] = 3\left[-10 + 10\right] = 0$$

The terms are $-5, -3, -1, 1, 3, 5$, which cancel in pairs — a useful sanity check that a sum of zero is not a mistake.

Question 16

The sum of the first $16$ terms of the AP $10$, $6$, $2$, … is

Solution. Here $a = 10$ and $d = -4$.

$$S_{16} = \frac{16}{2}\left[2(10) + 15(-4)\right] = 8\left[20 – 60\right] = 8(-40) = -320$$

The AP turns negative from the fourth term onward, so a negative sum is expected — which discards (B) immediately.

Question 17

In an AP if $a = 1$, $a_n = 20$ and $S_n = 399$, then $n$ is

Solution. Both the first and last terms are given, so use the second form of the sum formula — there is no need to find $d$ at all.

$$S_n = \frac{n}{2}\left(a + a_n\right) \implies 399 = \frac{n}{2}(1 + 20) = \frac{21n}{2}$$

$$n = \frac{798}{21} = 38$$

Question 18

The sum of the first five multiples of $3$ is

Solution. The first five multiples of $3$ are $3$, $6$, $9$, $12$, $15$ — an AP with $a = 3$, $d = 3$, $n = 5$.

$$S_5 = \frac{5}{2}(3 + 15) = \frac{5}{2}(18) = 45$$

Or simply $3(1+2+3+4+5) = 3 \times 15 = 45$.

The word “first” matters: the multiples start at $3$, not at $0$. Counting $0$ as the first multiple would give the same sum here, but it changes the answer in questions where the count is larger.

Common mistakes

• Using $n$ instead of $n-1$. $a_{21} = a + 20d$, not $a + 21d$. Option (C) in Question 6 is placed for exactly this, and it is the error that costs most marks in the whole chapter.
• Subtracting in the wrong order for $d$. The common difference is (later term) minus (earlier term), so for $-10, -6, -2$ it is $+4$. Question 3 offers $-4$ as an option.
• Counting terms instead of gaps. $a_{18} – a_{13} = 5d$, because there are five steps between the 13th and 18th terms. Option (D) in Question 9 is $6d$.
• Miscounting from the end in Question 13. The last term is the first from the end, so the fourth from the end is $l – 3d$.
• Assuming $d \ne 0$. Question 2 has $d = 0$, and a constant sequence is still an AP. Nothing in the definition forbids it.
• Finding $d$ when it is not needed in Question 17. With $a$, $a_n$ and $S_n$ given, $S_n = \frac{n}{2}(a + a_n)$ has one unknown. Going via $d$ needs two equations and several more lines.
• Trying to find $a$ and $d$ separately in Questions 11 and 12. In both, the unknowns cancel. Question 12 in particular looks under-determined — two unknowns, one equation — until you notice that it is the ratio $a : d$ that the condition fixes, and that is all the 18th term needs.

Practise next

• Exemplar Exercise 5.2 — deciding whether a list is an AP at all, and justifying it, which is the reasoning behind Questions 3 and 5 above.
• Exemplar Exercise 5.3 Q1–18 for the same $n$-th term work written out, and Q19–35 for the sums.
• NCERT textbook Exercise 5.2 — the $n$-th term formula on the book’s own problems.
• NCERT textbook Exercise 5.3 for $S_n$, including the Gauss pairing of Question 14.