NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions — Complete Guide with Step-by-Step Solutions

An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed number (the common difference) to the preceding term.

If the first term is $a$ and the common difference is $d$, the AP is:

Common difference: $d = a_2 - a_1 = a_3 - a_2 = \ldots = a_{n+1} - a_n$ (constant for all consecutive pairs).

The common difference can be positive (increasing AP: $2, 5, 8, 11, \ldots$), negative (decreasing AP: $10, 7, 4, 1, \ldots$), or zero (constant AP: $5, 5, 5, 5, \ldots$).

The $n$th term (also called the general term) of an AP is:

where $a$ = first term, $d$ = common difference, $n$ = position of the term.

This formula answers: "What is the 50th term?" or "Which term equals 100?" or "Is 301 a term of this AP?"

The last term is often denoted by $l$. If there are $n$ terms, $l = a + (n-1)d$.

The sum of the first $n$ terms of an AP is:

Alternative form (when the last term $l$ is known):

The second form is often faster and simpler. It says: the sum equals the number of terms times the average of the first and last terms.

Important relationship: $a_n = S_n - S_{n-1}$ (the $n$th term equals the difference of consecutive partial sums).

These properties appear frequently in problems:

1. If $a$, $b$, $c$ are in AP, then $2b = a + c$ (the middle term is the average of the other two).

2. If three numbers are in AP, take them as $a - d$, $a$, $a + d$.

3. If four numbers are in AP, take them as $a - 3d$, $a - d$, $a + d$, $a + 3d$.

4. $S_n = \frac{n}{2}(a + l)$ is especially useful when both the first and last terms are known.

5. Sum of first $n$ natural numbers: $S = \frac{n(n+1)}{2}$.

(i) $2, 4, 8, 16, \ldots$
$4 - 2 = 2$, $8 - 4 = 4$, $16 - 8 = 8$. Differences are not constant. Not an AP (it's a GP).

(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$
$\frac{5}{2} - 2 = \frac{1}{2}$, $3 - \frac{5}{2} = \frac{1}{2}$, $\frac{7}{2} - 3 = \frac{1}{2}$. Common difference $= \frac{1}{2}$. Yes, AP.
Next three: $4, \frac{9}{2}, 5$.

(iii) $-1.2, -3.2, -5.2, -7.2, \ldots$
$-3.2 - (-1.2) = -2$, $-5.2 - (-3.2) = -2$. Common difference $= -2$. Yes, AP.
Next three: $-9.2, -11.2, -13.2$.

(iv) $-10, -6, -2, 2, \ldots$
$-6 - (-10) = 4$, $-2 - (-6) = 4$. Common difference $= 4$. Yes, AP.
Next three: $6, 10, 14$.

(v) $3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \ldots$
Difference $= \sqrt{2}$ throughout. Yes, AP with $d = \sqrt{2}$.
Next three: $3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}$.

(vi) $0.2, 0.22, 0.222, 0.2222, \ldots$
$0.22 - 0.2 = 0.02$, $0.222 - 0.22 = 0.002$. Not constant. Not an AP.

(vii) $0, -4, -8, -12, \ldots$
Difference $= -4$ throughout. Yes, AP.
Next three: $-16, -20, -24$.

(viii) $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots$
Difference $= 0$. Yes, AP with $d = 0$. Every term is $-\frac{1}{2}$.

(ix) $1, 3, 9, 27, \ldots$
$3 - 1 = 2$, $9 - 3 = 6$. Not constant. Not an AP (it's a GP with ratio 3).

(x) $a, 2a, 3a, 4a, \ldots$
Difference $= a$ throughout. Yes, AP with $d = a$.
Next three: $5a, 6a, 7a$.

(i) $a = 10, d = 10$: $10, 20, 30, 40$
(ii) $a = -2, d = 0$: $-2, -2, -2, -2$
(iii) $a = 4, d = -3$: $4, 1, -2, -5$
(iv) $a = -1, d = \frac{1}{2}$: $-1, -\frac{1}{2}, 0, \frac{1}{2}$
(v) $a = -1.25, d = -0.25$: $-1.25, -1.50, -1.75, -2.00$

Solution:
$a = 3$, $d = 5$. We need $a_n = 78$.

$a_n = a + (n-1)d$
$78 = 3 + (n-1)(5)$
$75 = 5(n-1)$
$n - 1 = 15$
$n = 16$

Answer: $78$ is the $16$th term.

Solution:
$a = 11$, $d = -3$. Check if $a_n = -150$ for some positive integer $n$.

$-150 = 11 + (n-1)(-3)$
$-161 = -3(n-1)$
$n - 1 = \frac{161}{3} = 53.\overline{6}$

Since $n$ is not a whole number, $-150$ is not a term of this AP.

Solution:
$a = 10$, $d = 7 - 10 = -3$.
$a_{30} = 10 + (30-1)(-3) = 10 - 87 = -77$

Answer: The 30th term is $-77$.

Solution:
$a = -3$, $d = -\frac{1}{2} - (-3) = \frac{5}{2}$.
$a_{11} = -3 + 10 \times \frac{5}{2} = -3 + 25 = 22$

Answer: The 11th term is $22$.

Solution:
$a = 21$, $d = -3$.
$-81 = 21 + (n-1)(-3)$
$-102 = -3(n-1)$
$n - 1 = 34 \implies n = 35$

Answer: $-81$ is the $35$th term.

Solution:
$a = 7$, $d = 6$, $a_n = 205$.
$205 = 7 + (n-1)(6)$
$198 = 6(n-1)$
$n = 34$

Answer: There are $34$ terms.

Solution:
$a_{11} = a + 10d = 38 \quad \cdots (1)$
$a_{16} = a + 15d = 73 \quad \cdots (2)$

Subtract (1) from (2): $5d = 35 \implies d = 7$.
From (1): $a = 38 - 70 = -32$.

$a_{31} = -32 + 30(7) = -32 + 210 = 178$

Answer: The 31st term is $178$.

Solution:
$a = 3$, $d = 12$.
$a_{54} = 3 + 53(12) = 3 + 636 = 639$.

We need $a_n = 639 + 132 = 771$.
$771 = 3 + (n-1)(12)$
$768 = 12(n-1)$
$n = 65$

Answer: The $65$th term.

Solution:
$a_3 = a + 2d = 4 \quad \cdots (1)$
$a_9 = a + 8d = -8 \quad \cdots (2)$

Subtract: $6d = -12 \implies d = -2$. From (1): $a = 4 + 4 = 8$.

For $a_n = 0$: $8 + (n-1)(-2) = 0 \implies 8 - 2n + 2 = 0 \implies n = 5$.

Answer: The $5$th term is $0$.

Solution:
$a_{17} = 2a_8 + 5$:
$a + 16d = 2(a + 7d) + 5 = 2a + 14d + 5$
$-a + 2d = 5 \implies a = 2d - 5 \quad \cdots (1)$

$a_{11} = a + 10d = 43 \quad \cdots (2)$

Substitute (1) in (2): $(2d - 5) + 10d = 43 \implies 12d = 48 \implies d = 4$.
$a = 8 - 5 = 3$.

The AP is: $3, 7, 11, 15, 19, \ldots$

Solution:
AP1: first term $= 8$, AP2: first term $= 3$. Same $d$.

$a_{30}^{(1)} - a_{30}^{(2)} = [8 + 29d] - [3 + 29d] = 5$

Answer: The difference is $5$.

Key insight: The difference between corresponding terms of two APs with the same common difference is always constant (equal to the difference of their first terms).

Solution:
$a = 8$, $d = -5$, $n = 22$.

$S_{22} = \frac{22}{2}[2(8) + 21(-5)] = 11[16 - 105] = 11 \times (-89) = -979$

Answer: $S_{22} = -979$

Solution:
$a_n = 3 + 2n$. So $a_1 = 5$ and $a_{24} = 3 + 48 = 51$.

$S_{24} = \frac{24}{2}(a_1 + a_{24}) = 12(5 + 51) = 12 \times 56 = 672$

Answer: $S_{24} = 672$

Solution:
$a = 9$, $d = 8$.

$S_n = \frac{n}{2}[2(9) + (n-1)(8)] = \frac{n}{2}[18 + 8n - 8] = \frac{n}{2}(10 + 8n) = n(5 + 4n) = 636$

$4n^2 + 5n - 636 = 0$

Using the quadratic formula: $D = 25 + 4 \times 4 \times 636 = 25 + 10176 = 10201 = 101^2$.

$n = \frac{-5 + 101}{8} = \frac{96}{8} = 12$

Answer: $12$ terms are needed.

Solution:
The AP is $8, 16, 24, \ldots$ with $a = 8$, $d = 8$, $n = 15$.

$a_{15} = 8 + 14(8) = 120$.

$S_{15} = \frac{15}{2}(8 + 120) = \frac{15}{2} \times 128 = 15 \times 64 = 960$

Answer: $S_{15} = 960$

Solution:
Odd numbers: $1, 3, 5, 7, \ldots, 49$.
$a = 1$, $d = 2$, $l = 49$.

Number of terms: $49 = 1 + (n-1)(2) \implies n = 25$.

$S_{25} = \frac{25}{2}(1 + 49) = \frac{25}{2} \times 50 = 625$

Answer: Sum $= 625$.

Alternative: Sum of first $n$ odd numbers $= n^2$. So sum of first 25 odd numbers $= 25^2 = 625$. $\checkmark$

Solution:
$a = 17$, $l = 350$, $d = 9$.

$350 = 17 + (n-1)(9) \implies 333 = 9(n-1) \implies n = 38$.

$S_{38} = \frac{38}{2}(17 + 350) = 19 \times 367 = 6973$

Answer: $n = 38$, $S_{38} = 6973$.

Solution:
$d = a_3 - a_2 = 18 - 14 = 4$. $a = a_2 - d = 14 - 4 = 10$.

$S_{51} = \frac{51}{2}[2(10) + 50(4)] = \frac{51}{2}[20 + 200] = \frac{51}{2} \times 220 = 51 \times 110 = 5610$

Answer: $S_{51} = 5610$.

Problem: Subba Rao started work in 1995 at Rs 5000/month with an annual increment of Rs 200. In which year did his income reach Rs 7000?

Solution:
Monthly income in successive years: $5000, 5200, 5400, \ldots$ (AP with $a = 5000$, $d = 200$).

For income $= 7000$:
$7000 = 5000 + (n-1)(200)$
$2000 = 200(n-1)$
$n = 11$

Answer: In the $11$th year, i.e., $1995 + 10 = 2005$.

Problem: Find the sum of all natural numbers between 1 and 200 that are divisible by 5.

Solution:
Numbers: $5, 10, 15, \ldots, 200$. This is an AP with $a = 5$, $d = 5$, $l = 200$.

$n = \frac{200 - 5}{5} + 1 = 40$.

$S_{40} = \frac{40}{2}(5 + 200) = 20 \times 205 = 4100$

Answer: Sum $= 4100$.

Solution:
$a_3 = a + 2d = 7 \quad \cdots (1)$
$a_7 = a + 6d = 23 \quad \cdots (2)$

Subtract: $4d = 16 \implies d = 4$. From (1): $a = -1$.

$S_{10} = \frac{10}{2}[2(-1) + 9(4)] = 5[-2 + 36] = 5 \times 34 = 170$

Answer: $S_{10} = 170$.

Solution:
$S_n = 5n^2 + 3n$.

$a_n = S_n - S_{n-1} = [5n^2 + 3n] - [5(n-1)^2 + 3(n-1)]$
$= 5n^2 + 3n - 5n^2 + 10n - 5 - 3n + 3$
$= 10n - 2$

So $a_n = 10n - 2$.
$a_1 = 8$, $a_2 = 18$, $a_3 = 28$.

The AP is: $8, 18, 28, 38, \ldots$ with $d = 10$.

Verification: $S_3 = 5(9) + 3(3) = 45 + 9 = 54$. Sum of first 3 terms $= 8 + 18 + 28 = 54$ $\checkmark$.

Solution:
Let the two APs have first terms $a_1, a_2$ and common differences $d_1, d_2$.

$\frac{S_n^{(1)}}{S_n^{(2)}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n + 1}{4n + 27}$

We need the ratio of 11th terms: $\frac{a_1 + 10d_1}{a_2 + 10d_2}$.

Notice that $a + 10d = \frac{2a + 20d}{2}$. Setting $n - 1 = 20$, i.e., $n = 21$:

$\frac{2a_1 + 20d_1}{2a_2 + 20d_2} = \frac{7(21) + 1}{4(21) + 27} = \frac{148}{111} = \frac{4}{3}$

Answer: The ratio of 11th terms is $4:3$.

Solution:
$a_n = S_n - S_{n-1} = [3n^2 + 5n] - [3(n-1)^2 + 5(n-1)]$
$= 3n^2 + 5n - 3n^2 + 6n - 3 - 5n + 5 = 6n + 2$

Since $a_n = 6n + 2$ is a linear function of $n$, yes, it is an AP with $d = 6$.

$a_{20} = 6(20) + 2 = 122$.

Answer: Yes, it is an AP. $a_{20} = 122$.

Problem: Ramkali saved Rs 5 in the first week, Rs 6.50 in the second, Rs 8 in the third, and so on. Find her total savings after 52 weeks.

Solution:
$a = 5$, $d = 1.5$, $n = 52$.

$S_{52} = \frac{52}{2}[2(5) + 51(1.5)] = 26[10 + 76.5] = 26 \times 86.5 = 2249$

Answer: Total savings $=$ Rs $2249$.

Problem: 200 logs are stacked with 20 in the bottom row, 19 in the next, 18 in the next, etc. How many rows? How many logs in the top row?

Solution:
This is an AP: $20, 19, 18, \ldots$ with $a = 20$, $d = -1$.

$S_n = \frac{n}{2}[2(20) + (n-1)(-1)] = \frac{n}{2}[41 - n] = 200$

$n(41 - n) = 400$
$n^2 - 41n + 400 = 0$
$(n - 16)(n - 25) = 0$

$n = 16$ (if $n = 25$, the top row would have $20 - 24 = -4$ logs, impossible).

Top row: $a_{16} = 20 + 15(-1) = 5$ logs.

Answer: $16$ rows, $5$ logs in the top row.

Problem: A car travels 10 m in the first second, 15 m in the second, 20 m in the third, etc. How far does it travel in 20 seconds?

Solution:
$a = 10$, $d = 5$, $n = 20$.

$S_{20} = \frac{20}{2}[2(10) + 19(5)] = 10[20 + 95] = 10 \times 115 = 1150$ m.

Answer: $1150$ m.

Solution:
Two-digit multiples of 3: $12, 15, 18, \ldots, 99$.
$a = 12$, $d = 3$, $l = 99$.
$n = \frac{99 - 12}{3} + 1 = 30$.
$S_{30} = \frac{30}{2}(12 + 99) = 15 \times 111 = 1665$.

Answer: $1665$.

Solution:
$a_{25} = S_{25} - S_{24} = [3(625) + 100] - [3(576) + 96]$
$= [1875 + 100] - [1728 + 96] = 1975 - 1824 = 151$.

Or use $a_n = 6n + 1$: $a_{25} = 150 + 1 = 151$.

Answer: $a_{25} = 151$.

Solution:
$S_{10} = 4(10) - 100 = 40 - 100 = -60$.
$a_{10} = S_{10} - S_9 = -60 - [4(9) - 81] = -60 - (36 - 81) = -60 + 45 = -15$.

Alternatively: $a_n = S_n - S_{n-1} = (4n - n^2) - (4(n-1) - (n-1)^2) = 5 - 2n$.
$a_{10} = 5 - 20 = -15$. $\checkmark$

Answer: $S_{10} = -60$, $a_{10} = -15$.