NCERT Exemplar Solutions Class 10 Maths Arithmetic Progressions — Exercise 5.3 Q1–18

NCERT Exemplar Problems, Class 10 Mathematics — Arithmetic Progressions, Exercise 5.3, questions 1 to 18.

Exercise 5.3 is thirty-five questions long, so it is split across two pages here; this one covers the first eighteen, which are all about terms, and the second covers questions 19 to 35, which are about sums.

Almost every question here reduces to the same move: turn the two facts you are given into two equations in $a$ and $d$, and solve. The variations are worth naming:

• Symmetric labelling. When three numbers are in AP and their sum is known, write them as $a – d$, $a$, $a + d$ — the $d$s cancel and $a$ falls out immediately. Questions 12 and 13 both do this.
• Differences of terms. $a_p – a_q = (p-q)d$ turns a two-term condition into a value of $d$ with no algebra at all. Questions 5, 6 and 8 use it.
• “Is $x$ a term?” Solve $a + (n-1)d = x$ and check that $n$ is a positive integer. Question 10 is a yes; Question 5 of Exercise 5.2 was a no.

Key insight. An AP is determined by exactly two numbers, $a$ and $d$. So any question that gives you two independent facts about it has enough information — and any question that seems to give only one is really giving you a ratio, as Question 11 does.

Question 1

Match the APs given in column A with suitable common differences given in column B.

Column A		Column B
$(A_1)$	$2$, $-2$, $-6$, $-10$, …	$(B_1)$	$\dfrac23$
$(A_2)$	$a = -18$, $n = 10$, $a_n = 0$	$(B_2)$	$-5$
$(A_3)$	$a = 0$, $a_{10} = 6$	$(B_3)$	$4$
$(A_4)$	$a_2 = 13$, $a_4 = 3$	$(B_4)$	$-4$
		$(B_5)$	$2$
		$(B_6)$	$\dfrac12$
		$(B_7)$	$5$

Solution. Four separate one-line calculations.

$(A_1)$ Read the difference straight off: $-2 – 2 = -4$. → $(B_4)$

$(A_2)$ Use $a_n = a + (n-1)d$ with $n = 10$:

$$0 = -18 + 9d \implies d = 2$$ → $(B_5)$

$(A_3)$ $a_{10} = a + 9d$, so $6 = 0 + 9d$ and $d = \dfrac69 = \dfrac23$ → $(B_1)$

$(A_4)$ Two terms two places apart, so $a_4 – a_2 = 2d$:

$$3 – 13 = 2d \implies d = -5$$ → $(B_2)$

Note that column B has seven entries for four APs — three are there only as distractors, so there is no matching-by-elimination shortcut.

Question 2

Verify that each of the following is an AP, and then write its next three terms.

(i) $0$, $\dfrac14$, $\dfrac12$, $\dfrac34$, …

Differences: $\frac14$, $\frac14$, $\frac14$ — constant, so it is an AP with $d = \frac14$. Continuing:

$$1, \quad \frac54, \quad \frac32$$

(ii) $5$, $\dfrac{14}{3}$, $\dfrac{13}{3}$, $4$, …

Write $5 = \frac{15}{3}$ and $4 = \frac{12}{3}$ to see the pattern: $\frac{15}{3}, \frac{14}{3}, \frac{13}{3}, \frac{12}{3}$. The difference is $-\frac13$ throughout. Next three:

$$\frac{11}{3}, \quad \frac{10}{3}, \quad 3$$

(iii) $\sqrt3$, $2\sqrt3$, $3\sqrt3$, …

Difference $\sqrt3$ each time. Next three:

$$4\sqrt3, \quad 5\sqrt3, \quad 6\sqrt3$$

(iv) $a + b$, $(a+1) + b$, $(a+1) + (b+1)$, …

Simplify each term: $a + b$, $a + b + 1$, $a + b + 2$. The difference is $1$, so it is an AP. The pattern of the printed form alternates which letter is increased, so the next three keep alternating:

$$(a+2) + (b+1), \quad (a+2) + (b+2), \quad (a+3) + (b+2)$$

which are $a+b+3$, $a+b+4$, $a+b+5$.

(v) $a$, $2a + 1$, $3a + 2$, $4a + 3$, …

Differences: $(2a+1) – a = a + 1$; $(3a+2) – (2a+1) = a+1$; $(4a+3)-(3a+2) = a+1$. Constant, so it is an AP with $d = a + 1$. Next three:

$$5a + 4, \quad 6a + 5, \quad 7a + 6$$

Question 3

Write the first three terms of the APs when $a$ and $d$ are as given below.

(i) $a = \dfrac12$, $d = -\dfrac16$

$$\frac12, \quad \frac12 – \frac16 = \frac13, \quad \frac13 – \frac16 = \frac16$$

(ii) $a = -5$, $d = -3$

$$-5, \quad -8, \quad -11$$

(iii) $a = \sqrt2$, $d = \dfrac{1}{\sqrt2}$

$$\sqrt2, \quad \sqrt2 + \frac{1}{\sqrt2} = \frac{3}{\sqrt2}, \quad \frac{3}{\sqrt2} + \frac{1}{\sqrt2} = \frac{4}{\sqrt2}$$

using $\sqrt2 = \frac{2}{\sqrt2}$ for the middle step.

Question 4

Find $a$, $b$ and $c$ such that the following numbers are in AP: $a$, $7$, $b$, $23$, $c$.

Solution. The two known terms, $7$ and $23$, are two places apart, so

$$23 – 7 = 2d \implies d = 8$$

Now step outward and inward from them:

$$a = 7 – d = -1, \qquad b = 7 + d = 15, \qquad c = 23 + d = 31$$

Check the whole list: $-1$, $7$, $15$, $23$, $31$ — differences all $8$ ✓.

Question 5

Determine the AP whose fifth term is $19$ and the difference of the eighth term from the thirteenth term is $20$.

Solution. The second condition is a difference of terms, so it gives $d$ at once:

$$a_{13} – a_8 = 5d = 20 \implies d = 4$$

Then from $a_5 = 19$:

$$a + 4(4) = 19 \implies a = 3$$

The AP is $3$, $7$, $11$, $15$, …

Question 6

The $26$th, $11$th and the last term of an AP are $0$, $3$ and $-\dfrac15$ respectively. Find the common difference and the number of terms.

Solution. Use the two terms whose positions are known:

$$a_{11} – a_{26} = (11 – 26)d = -15d = 3 – 0 = 3 \implies d = -\frac15$$

From $a_{26} = 0$:

$$a + 25\left(-\frac15\right) = 0 \implies a = 5$$

Now find which term equals $-\frac15$:

$$5 + (n-1)\left(-\frac15\right) = -\frac15 \implies (n-1)\frac15 = 5 + \frac15 = \frac{26}{5} \implies n – 1 = 26$$

so $n = 27$.

Question 7

The sum of the $5$th and the $7$th terms of an AP is $52$ and the $10$th term is $46$. Find the AP.

Solution. Two equations in $a$ and $d$:

$$(a + 4d) + (a + 6d) = 52 \implies 2a + 10d = 52 \implies a + 5d = 26$$ $$a + 9d = 46$$

Subtracting, $4d = 20$, so $d = 5$ and $a = 26 – 25 = 1$.

The AP is $1$, $6$, $11$, $16$, …

Notice that $a + 5d$ is the sixth term, so the first condition says $a_6 = 26$ — the average of the 5th and 7th terms is the 6th, which is true in any AP and is a quick way to halve the work.

Question 8

Find the $20$th term of the AP whose $7$th term is $24$ less than the $11$th term, first term being $12$.

Solution. “The 7th term is 24 less than the 11th” means $a_{11} – a_7 = 24$, so

$$4d = 24 \implies d = 6$$

With $a = 12$,

$$a_{20} = 12 + 19(6) = 12 + 114 = 126$$

Question 9

If the $9$th term of an AP is zero, prove that its $29$th term is twice its $19$th term.

Solution. The condition $a_9 = 0$ gives

$$a + 8d = 0 \implies a = -8d$$

Now express both terms in $d$ alone:

$$a_{29} = a + 28d = -8d + 28d = 20d$$ $$a_{19} = a + 18d = -8d + 18d = 10d$$

Hence $a_{29} = 20d = 2 \times 10d = 2a_{19}$, which is what was to be proved.

The structure is worth seeing: a single condition on an AP fixes the ratio of $a$ to $d$, and any relation between terms that is homogeneous — as “one term is twice another” is — then follows.

Question 10

Find whether $55$ is a term of the AP $7$, $10$, $13$, … or not. If yes, find which term it is.

Solution. Here $a = 7$, $d = 3$. Set the $n$-th term to $55$:

$$7 + (n-1)3 = 55 \implies 3(n-1) = 48 \implies n – 1 = 16 \implies n = 17$$

Since $17$ is a positive integer, $55$ is a term — the 17th.

Question 11

Determine $k$ so that $k^2 + 4k + 8$, $2k^2 + 3k + 6$, $3k^2 + 4k + 4$ are three consecutive terms of an AP.

Solution. Three terms are consecutive terms of an AP exactly when the middle one is the average of the outer two — equivalently, when the two differences agree:

$$\left(2k^2 + 3k + 6\right) – \left(k^2 + 4k + 8\right) = \left(3k^2 + 4k + 4\right) – \left(2k^2 + 3k + 6\right)$$

$$k^2 – k – 2 = k^2 + k – 2$$

The $k^2$ and $-2$ cancel on both sides, leaving

$$-k = k \implies 2k = 0 \implies k = 0$$

Check: with $k = 0$ the three terms are $8$, $6$, $4$ — an AP with $d = -2$ ✓.

Question 12

Split $207$ into three parts such that these are in AP and the product of the two smaller parts is $4623$.

Solution. Take the three parts as $a – d$, $a$, $a + d$ — the symmetric labelling, which makes the sum trivial:

$$(a – d) + a + (a + d) = 3a = 207 \implies a = 69$$

The two smaller parts are $a – d$ and $a$ (assuming $d > 0$), so

$$69(69 – d) = 4623 \implies 69 – d = 67 \implies d = 2$$

The parts are $67$, $69$, $71$. Check: $67 + 69 + 71 = 207$ ✓ and $67 \times 69 = 4623$ ✓.

Question 13

The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

Solution. Again take the angles as $a – d$, $a$, $a + d$ with $d > 0$, so the least is $a – d$ and the greatest is $a + d$.

Angle sum: $\ 3a = 180 \implies a = 60$.

Greatest is twice the least:

$$a + d = 2(a – d) \implies a + d = 2a – 2d \implies 3d = a = 60 \implies d = 20$$

The angles are $40^\circ$, $60^\circ$, $80^\circ$. Check: they sum to $180^\circ$, they are in AP with $d = 20$, and $80 = 2 \times 40$ ✓.

In any triangle whose angles are in AP the middle angle is $60^\circ$, because the angle sum is three times it — worth remembering.

Question 14

If the $n$th terms of the two APs $9$, $7$, $5$, … and $24$, $21$, $18$, … are the same, find the value of $n$. Also find that term.

Solution. The first AP has $a = 9$, $d = -2$; the second has $a = 24$, $d = -3$. Setting the $n$-th terms equal:

$$9 + (n-1)(-2) = 24 + (n-1)(-3)$$ $$9 – 2n + 2 = 24 – 3n + 3$$ $$11 – 2n = 27 – 3n \implies n = 16$$

The common term is

$$9 + 15(-2) = 9 – 30 = -21$$

(and from the second AP, $24 + 15(-3) = 24 – 45 = -21$ ✓).

Question 15

If the sum of the $3$rd and the $8$th terms of an AP is $7$ and the sum of the $7$th and the $14$th terms is $-3$, find the $10$th term.

Solution.

$$(a + 2d) + (a + 7d) = 7 \implies 2a + 9d = 7$$ $$(a + 6d) + (a + 13d) = -3 \implies 2a + 19d = -3$$

Subtracting, $10d = -10$, so $d = -1$ and $2a = 7 + 9 = 16$, giving $a = 8$.

$$a_{10} = 8 + 9(-1) = -1$$

Question 16

Find the $12$th term from the end of the AP $-2$, $-4$, $-6$, …, $-100$.

Solution. Counting from the end avoids finding how many terms there are. With last term $l = -100$ and $d = -2$, the $k$-th term from the end is $l – (k-1)d$:

$$\text{12th from the end} = -100 – 11(-2) = -100 + 22 = -78$$

(For confirmation: the AP is the even numbers from $-2$ to $-100$, so it has $50$ terms, and the 12th from the end is the $39$th from the start: $-2 + 38(-2) = -78$ ✓.)

Question 17

Which term of the AP $53$, $48$, $43$, … is the first negative term?

Solution. Here $a = 53$ and $d = -5$, so

$$a_n = 53 + (n-1)(-5) = 58 – 5n$$

The first negative term is the smallest $n$ with $a_n < 0$:

$$58 – 5n < 0 \implies n > \frac{58}{5} = 11.6$$

The smallest integer greater than $11.6$ is $12$, so the 12th term is the first negative one. Indeed $a_{11} = 3$ and $a_{12} = -2$.

The strict inequality matters: an inequality question asks for the first $n$ that crosses the boundary, not the $n$ that lands on it.

Question 18

How many numbers lie between $10$ and $300$, which when divided by $4$ leave a remainder $3$?

Solution. Numbers leaving remainder $3$ on division by $4$ are $\dots, 7, 11, 15, 19, \dots$ — an AP with $d = 4$.

The first one after 10 is $11$; the last one before 300 is $299$ (since $299 = 4 \times 74 + 3$, and the next, $303$, is past $300$). So the list is

$$11, \ 15, \ 19, \ \dots, \ 299$$

Counting it:

$$299 = 11 + (n-1)4 \implies 4(n-1) = 288 \implies n – 1 = 72 \implies n = 73$$

Common mistakes

• Not using the symmetric form in Questions 12 and 13. Writing the three parts as $a$, $a+d$, $a+2d$ works but leaves $3a + 3d = 207$, one more term to carry through every line. The $a-d$, $a$, $a+d$ form makes the sum give $a$ immediately.
• Assuming $d > 0$ without saying so. In Question 12 “the two smaller parts” and in Question 13 “the greatest” and “the least” both presuppose an ordering. Stating it costs one clause and makes the working correct rather than lucky.
• Using $\le$ instead of $<$ in Question 17. The first negative term needs $a_n < 0$ strictly; $a_n \le 0$ would admit a zero term, which is neither positive nor negative.
• Getting the endpoints wrong in Question 18. “Between 10 and 300” starts at $11$, not $7$, and ends at $299$, not $303$. Writing out the first and last candidate explicitly is the cheapest guard.
• Counting terms rather than gaps in Question 16. The last term is the first from the end, so the 12th from the end is $l – 11d$.
• Expanding $k^2$ terms carefully in Question 11. Both sides carry $k^2$ and both carry $-2$; once those cancel the equation is linear, and a sign slip in the expansion is the only way to get a different answer.
• Solving for $a$ before $d$. In Questions 5, 6 and 8 the difference of two terms gives $d$ in one line with no reference to $a$ at all. Reach for $a_p – a_q = (p-q)d$ first, every time.

Practise next

• Exemplar Exercise 5.3 Q19–35 — the sums half of this exercise, which assumes everything above.
• Exemplar Exercise 5.4 for the long-answer versions, including the ratio and instalment problems.
• NCERT textbook Exercise 5.2 — twenty questions on the $n$-th term, the closest match to Questions 4 to 18 above.
• Exemplar Exercise 5.2 if the “is this really an AP?” step in Question 2 needs more work.