NCERT Class 10 Mathematics — Arithmetic Progressions, Exercise 5.2. All 20 questions solved.

Every question in Exercise 5.2 runs on one formula. The $n$th term of the AP with first term $a$ and common difference $d$ is

$$a_n = a + (n – 1)d.$$

It links four quantities, $a$, $d$, $n$ and $a_n$. A question gives you some of them and asks for the rest. Two consequences do much of the work:

• If two terms are known, writing each with the formula gives two linear equations in $a$ and $d$, which can be solved as a pair.
• The gap between two terms depends only on $d$: $a_m – a_k = (m – k)d$. The first term cancels, so questions about a difference of terms never need $a$.

Key insight. Each question here is a translation exercise into $a_n = a + (n – 1)d$. Decide which of $a$, $d$, $n$, $a_n$ the words give you, write down the equation (or pair of equations), and solve. For “which term…?” questions the answer must also be a positive whole number. If $n$ comes out as a fraction, the number is not in the list.

Question 1

Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP:

$$\begin{array}{c|c|c|c|c} & a & d & n & a_n \\ \hline \text{(i)} & 7 & 3 & 8 & \ldots \\ \text{(ii)} & -18 & \ldots & 10 & 0 \\ \text{(iii)} & \ldots & -3 & 18 & -5 \\ \text{(iv)} & -18.9 & 2.5 & \ldots & 3.6 \\ \text{(v)} & 3.5 & 0 & 105 & \ldots \end{array}$$

Solution. Each row gives three of the four quantities; substitute them into $a_n = a + (n – 1)d$ and solve for the fourth.

(i) $a_8 = 7 + 7 \times 3 = 28$.

(ii) $0 = -18 + 9d$, so $d = 2$.

(iii) $-5 = a + 17(-3) = a – 51$, so $a = 46$.

(iv) $3.6 = -18.9 + (n – 1)(2.5)$, so $(n – 1)(2.5) = 22.5$, $n – 1 = 9$ and $n = 10$.

(v) With $d = 0$ every term equals the first: $a_{105} = 3.5 + 104 \times 0 = 3.5$.

Question 2

Choose the correct choice in the following and justify:

(i) $30$th term of the AP: $10, 7, 4, \ldots$, is

(ii) $11$th term of the AP: $-3, -\dfrac{1}{2}, 2, \ldots$, is

Solution.

(i) Here $a = 10$ and $d = 7 – 10 = -3$, so

$$a_{30} = 10 + 29(-3) = 10 – 87 = -77$$

which is option (C). The list is falling by $3$ each time, so a large positive answer such as (A) or (B) was never possible.

(ii) Here $a = -3$ and $d = -\dfrac{1}{2} – (-3) = \dfrac{5}{2}$. The list is rising, since $-3, -\frac{1}{2}, 2$ go up. So

$$a_{11} = -3 + 10 \times \frac{5}{2} = -3 + 25 = 22$$

which is option (B).

Question 3

In the following APs, find the missing terms in the boxes:

(i) $2,\ \square,\ 26$    (ii) $\square,\ 13,\ \square,\ 3$    (iii) $5,\ \square,\ \square,\ 9\frac{1}{2}$

(iv) $-4,\ \square,\ \square,\ \square,\ \square,\ 6$    (v) $\square,\ 38,\ \square,\ \square,\ \square,\ -22$

Solution. Count positions carefully: the known numbers fix two terms, and the gap between them, divided by the number of steps, gives $d$.

(i) $a_1 = 2$ and $a_3 = 26$ are two steps apart, so $2d = 24$ and $d = 12$. The missing term is $2 + 12 = 14$. (The middle of three terms in an AP is always the average of the outer two.)

(ii) Here $13$ is the second term and $3$ the fourth: $2d = 3 – 13 = -10$, so $d = -5$. Then $a_1 = 13 + 5 = 18$ and $a_3 = 13 – 5 = 8$.

(iii) $a_1 = 5$ and $a_4 = 9\frac{1}{2}$ are three steps apart: $3d = \frac{9}{2}$, so $d = \frac{3}{2}$. The missing terms are $6\frac{1}{2}$ and $8$.

(iv) $a_1 = -4$ and $a_6 = 6$ are five steps apart: $5d = 10$, $d = 2$. The missing terms are $-2, 0, 2, 4$.

(v) $a_2 = 38$ and $a_6 = -22$ are four steps apart: $4d = -60$, $d = -15$. So $a_1 = 38 + 15 = 53$, and the terms after $38$ are $23, 8, -7$ before reaching $-22$.

Question 4

Which term of the AP: $3, 8, 13, 18, \ldots$, is $78$?

Solution. Here $a = 3$ and $d = 5$. We want $n$ with $a_n = 78$:

$$3 + (n – 1)5 = 78 \quad\Longrightarrow\quad 5(n – 1) = 75 \quad\Longrightarrow\quad n = 16$$

Question 5

Find the number of terms in each of the following APs:

(i) $7, 13, 19, \ldots, 205$    (ii) $18, 15\frac{1}{2}, 13, \ldots, -47$

Solution. The number of terms is the position $n$ of the last term.

(i) $a = 7$, $d = 6$, last term $205$:

$$7 + 6(n – 1) = 205 \quad\Longrightarrow\quad 6(n – 1) = 198 \quad\Longrightarrow\quad n – 1 = 33 \quad\Longrightarrow\quad n = 34$$

(ii) $a = 18$ and $d = 15\frac{1}{2} – 18 = -\frac{5}{2}$, last term $-47$:

$$18 – \frac{5}{2}(n – 1) = -47 \quad\Longrightarrow\quad \frac{5}{2}(n – 1) = 65 \quad\Longrightarrow\quad n – 1 = 26 \quad\Longrightarrow\quad n = 27$$

Question 6

Check whether $-150$ is a term of the AP: $11, 8, 5, 2, \ldots$

Solution. Here $a = 11$ and $d = -3$. If $-150$ were the $n$th term,

$$11 – 3(n – 1) = -150 \quad\Longrightarrow\quad 3(n – 1) = 161 \quad\Longrightarrow\quad n = \frac{164}{3}$$

A term’s position must be a positive whole number, and $\frac{164}{3} = 54\frac{2}{3}$ is not. The list steps from $a_{54} = -148$ straight to $a_{55} = -151$, missing $-150$.

Question 7

Find the $31$st term of an AP whose $11$th term is $38$ and the $16$th term is $73$.

Solution. Two known terms give two equations:

$$a + 10d = 38, \qquad a + 15d = 73$$

Subtracting, $5d = 35$, so $d = 7$, and then $a = 38 – 70 = -32$. Therefore

$$a_{31} = -32 + 30 \times 7 = 178$$

Question 8

An AP consists of $50$ terms of which $3$rd term is $12$ and the last term is $106$. Find the $29$th term.

Solution. “The last term” of a $50$-term AP is $a_{50}$. So

$$a + 2d = 12, \qquad a + 49d = 106$$

Subtracting, $47d = 94$, so $d = 2$ and $a = 12 – 4 = 8$. Then

$$a_{29} = 8 + 28 \times 2 = 64$$

Question 9

If the $3$rd and the $9$th terms of an AP are $4$ and $-8$ respectively, which term of this AP is zero?

Solution. From $a + 2d = 4$ and $a + 8d = -8$, subtracting gives $6d = -12$, so $d = -2$ and $a = 8$. For the zero term,

$$8 + (n – 1)(-2) = 0 \quad\Longrightarrow\quad n – 1 = 4 \quad\Longrightarrow\quad n = 5$$

Check: the AP is $8, 6, 4, 2, 0, \ldots$, and $0$ sits in fifth place.

Question 10

The $17$th term of an AP exceeds its $10$th term by $7$. Find the common difference.

Solution. The gap between two terms involves only $d$:

$$a_{17} – a_{10} = (a + 16d) – (a + 9d) = 7d = 7$$

so $d = 1$. The first term cancels, which is why it is not needed (and cannot be found).

Question 11

Which term of the AP: $3, 15, 27, 39, \ldots$ will be $132$ more than its $54$th term?

Solution. Here $a = 3$ and $d = 12$, so $a_{54} = 3 + 53 \times 12 = 639$. We need the term equal to $639 + 132 = 771$:

$$3 + 12(n – 1) = 771 \quad\Longrightarrow\quad 12(n – 1) = 768 \quad\Longrightarrow\quad n = 65$$

A quicker way to see it: $132 = 11 \times 12 = 11d$, so the required term is $11$ places after the $54$th, which is the $65$th.

Question 12

Two APs have the same common difference. The difference between their $100$th terms is $100$, what is the difference between their $1000$th terms?

Solution. Let the first terms be $a$ and $b$, and the shared common difference $d$. Then

$$a_{100} – b_{100} = (a + 99d) – (b + 99d) = a – b = 100$$

At any position $n$ the $(n – 1)d$ parts cancel in the same way, leaving $a – b$. So the two APs stay exactly $100$ apart for ever:

$$a_{1000} – b_{1000} = (a + 999d) – (b + 999d) = a – b = 100$$

Question 13

How many three-digit numbers are divisible by $7$?

Solution. The three-digit multiples of $7$ form an AP with $d = 7$. The smallest is $105 = 7 \times 15$ (since $98$ has only two digits), and the largest is $994 = 7 \times 142$ (since $1001$ has four). So

$$105 + 7(n – 1) = 994 \quad\Longrightarrow\quad 7(n – 1) = 889 \quad\Longrightarrow\quad n – 1 = 127 \quad\Longrightarrow\quad n = 128$$

Question 14

How many multiples of $4$ lie between $10$ and $250$?

Solution. The first multiple of $4$ after $10$ is $12$, and the last before $250$ is $248$. These and everything between form an AP with $d = 4$:

$$12 + 4(n – 1) = 248 \quad\Longrightarrow\quad 4(n – 1) = 236 \quad\Longrightarrow\quad n – 1 = 59 \quad\Longrightarrow\quad n = 60$$

Question 15

For what value of $n$, are the $n$th terms of two APs: $63, 65, 67, \ldots$ and $3, 10, 17, \ldots$ equal?

Solution. The first AP has $a = 63$, $d = 2$; the second has $a = 3$, $d = 7$. Setting their $n$th terms equal:

$$63 + 2(n – 1) = 3 + 7(n – 1) \quad\Longrightarrow\quad 60 = 5(n – 1) \quad\Longrightarrow\quad n = 13$$

Check: both $13$th terms are $87$. The second AP starts $60$ behind but gains $5$ per term, so it catches up after $12$ steps.

Question 16

Determine the AP whose third term is $16$ and the $7$th term exceeds the $5$th term by $12$.

Solution. The second condition is a difference of terms, so it gives $d$ alone: $a_7 – a_5 = 2d = 12$, so $d = 6$. The first condition then gives $a$: $a + 2(6) = 16$, so $a = 4$.

Question 17

Find the $20$th term from the last term of the AP: $3, 8, 13, \ldots, 253$.

Solution. Read the AP backwards. Starting from $253$ and stepping down by $5$ gives another AP, $253, 248, 243, \ldots$, with first term $253$ and $d = -5$. Its $20$th term is

$$253 + 19(-5) = 253 – 95 = 158$$

As a check from the front: $253 = 3 + 5(n – 1)$ gives $n = 51$ terms, and the $20$th from the end is the $(51 – 20 + 1) = 32$nd from the start, $3 + 31 \times 5 = 158$.

Question 18

The sum of the $4$th and $8$th terms of an AP is $24$ and the sum of the $6$th and $10$th terms is $44$. Find the first three terms of the AP.

Solution. Write each term with the formula:

$$(a + 3d) + (a + 7d) = 2a + 10d = 24, \qquad (a + 5d) + (a + 9d) = 2a + 14d = 44$$

Subtracting, $4d = 20$, so $d = 5$, and then $2a = 24 – 50 = -26$, so $a = -13$.

Question 19

Subba Rao started work in $1995$ at an annual salary of ₹ $5000$ and received an increment of ₹ $200$ each year. In which year did his income reach ₹ $7000$?

Solution. His salaries in successive years, $5000, 5200, 5400, \ldots$, form an AP with $a = 5000$ and $d = 200$. Find $n$ with $a_n = 7000$:

$$5000 + 200(n – 1) = 7000 \quad\Longrightarrow\quad 200(n – 1) = 2000 \quad\Longrightarrow\quad n = 11$$

So his salary reaches ₹ $7000$ in his $11$th year of work. Since $1995$ is the first year, the $11$th is $1995 + 10 = 2005$.

Question 20

Ramkali saved ₹ $5$ in the first week of a year and then increased her weekly savings by ₹ $1.75$. If in the $n$th week, her weekly savings become ₹ $20.75$, find $n$.

Solution. Her weekly savings form an AP with $a = 5$ and $d = 1.75$:

$$5 + 1.75(n – 1) = 20.75 \quad\Longrightarrow\quad 1.75(n – 1) = 15.75 \quad\Longrightarrow\quad n – 1 = 9 \quad\Longrightarrow\quad n = 10$$

Common mistakes

• Question 2(ii), getting the sign of $d$ wrong. $d = a_2 – a_1 = -\frac{1}{2} – (-3) = +\frac{5}{2}$. Subtracting the other way gives $d = -\frac{5}{2}$ and $a_{11} = -28$, which is not among the options. When your answer matches none of the choices, recheck the sign of $d$ first.
• Question 4, using $n$ in place of $n – 1$. The first term already is $a$, so reaching the $n$th term takes only $n – 1$ steps of $d$. Writing $3 + 5n = 78$ gives $n = 15$, one short.
• Question 3(ii) and (v), treating the first number shown as $a$. When the list opens with a box, the first visible number is the second term. Miscount the positions and every missing term comes out wrong.
• Question 6, rounding $n$. $n = 54\frac{2}{3}$ does not mean “about the 55th term”. A position has to be a whole number, and a fractional $n$ is the proof that $-150$ is not in the list.
• Questions 13 and 14, counting from the wrong end-points. In question 13 the AP starts at $105$, not at $100$ (which $7$ does not divide). In question 14 it starts at $12$ and ends at $248$. Forgetting the final “$+1$” when counting, $(994 – 105) \div 7 = 127$, is the other common slip.
• Question 17, counting 20 places from the end as the 31st term. From the front, the $20$th from the last of $51$ terms is the $32$nd, not the $31$st. Reversing the AP avoids the off-by-one.
• Question 19, answering “2006”. $1995$ is year $1$, so year $11$ is $1995 + 10 = 2005$. Adding $11$ counts one year too many.

Practise next

• Exercise 5.3 — the sum of the first $n$ terms, $S_n = \frac{n}{2}[2a + (n – 1)d]$, built on the same $a$, $d$ and $n$.
• Exercise 5.1 — revise recognising an AP and finding $d$, which every question here takes for granted.
• Class 11 Sequences and Series, Exercise 8.2 — geometric progressions, where the $n$th term becomes $ar^{n-1}$ and the same “two terms give two equations” method applies.