Arithmetic Progressions — NCERT Solutions

(i) Taxi fare situation:

Given: Fare for 1st km = ₹15, each additional km = ₹8.

Fare after 1st km = ₹15
Fare after 2nd km = ₹15 + ₹8 = ₹23
Fare after 3rd km = ₹23 + ₹8 = ₹31
Fare after 4th km = ₹31 + ₹8 = ₹39

The list is: 15, 23, 31, 39, …

Differences: $23 - 15 = 8$, $31 - 23 = 8$, $39 - 31 = 8$

Since each successive difference is the same (= 8), this list forms an AP with $a = 15$ and $d = 8$.

(ii) Air in cylinder:

Let initial amount of air = $V$.

After 1st pump: $V - \dfrac{V}{4} = \dfrac{3V}{4}$

After 2nd pump: $\dfrac{3V}{4} - \dfrac{1}{4}\cdot\dfrac{3V}{4} = \dfrac{3V}{4}\times\dfrac{3}{4} = \dfrac{9V}{16}$

After 3rd pump: $\dfrac{9V}{16}\times\dfrac{3}{4} = \dfrac{27V}{64}$

The list is: $V,\ \dfrac{3V}{4},\ \dfrac{9V}{16},\ \dfrac{27V}{64},\ldots$

Difference: $\dfrac{3V}{4} - V = -\dfrac{V}{4}$

$\dfrac{9V}{16} - \dfrac{3V}{4} = \dfrac{9V}{16} - \dfrac{12V}{16} = -\dfrac{3V}{16}$

Since $-\dfrac{V}{4} \neq -\dfrac{3V}{16}$, the differences are not equal.

This list does NOT form an AP.

(iii) Cost of digging a well:

Cost for 1st metre = ₹150
Cost for 2nd metre = ₹150 + ₹50 = ₹200
Cost for 3rd metre = ₹200 + ₹50 = ₹250
Cost for 4th metre = ₹250 + ₹50 = ₹300

The list is: 150, 200, 250, 300, …

Differences: $200-150=50$, $250-200=50$, $300-250=50$

Since each successive difference is the same (= 50), this list forms an AP with $a = 150$ and $d = 50$.

(iv) Compound interest:

Amount after 1st year = $10000\left(1+\dfrac{8}{100}\right)^1 = 10800$

Amount after 2nd year = $10000\left(1+\dfrac{8}{100}\right)^2 = 11664$

Amount after 3rd year = $10000\left(1+\dfrac{8}{100}\right)^3 = 12597.12$

Differences: $11664 - 10800 = 864$; $12597.12 - 11664 = 933.12$

Since the differences are not equal, this list does NOT form an AP.

The first four terms of an AP are: $a,\ a+d,\ a+2d,\ a+3d$.

(i) $a = 10$, $d = 10$

$$a_1 = 10,\quad a_2 = 10+10 = 20,\quad a_3 = 20+10 = 30,\quad a_4 = 30+10 = 40$$

First four terms: 10, 20, 30, 40

(ii) $a = 4$, $d = -3$

$$a_1 = 4,\quad a_2 = 4+(-3) = 1,\quad a_3 = 1+(-3) = -2,\quad a_4 = -2+(-3) = -5$$

First four terms: 4, 1, −2, −5

(iii) $a = -2$, $d = 0$

$$a_1 = -2,\quad a_2 = -2+0 = -2,\quad a_3 = -2,\quad a_4 = -2$$

First four terms: −2, −2, −2, −2

(iv) $a = -1$, $d = \dfrac{1}{2}$

$$a_1 = -1,\quad a_2 = -1+\frac{1}{2} = -\frac{1}{2},\quad a_3 = -\frac{1}{2}+\frac{1}{2} = 0,\quad a_4 = 0+\frac{1}{2} = \frac{1}{2}$$

First four terms: $-1,\ -\dfrac{1}{2},\ 0,\ \dfrac{1}{2}$

(v) $a = -1.25$, $d = -0.25$

$$a_1 = -1.25,\quad a_2 = -1.25+(-0.25) = -1.50,\quad a_3 = -1.75,\quad a_4 = -2.00$$

First four terms: −1.25, −1.50, −1.75, −2.00

(i) $3, 1, -1, -3, \ldots$

First term $a = 3$

Common difference $d = 1 - 3 = -2$

(ii) $-5, -1, 3, 7, \ldots$

First term $a = -5$

Common difference $d = -1 - (-5) = 4$

(iii) $\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{13}{3}, \ldots$

First term $a = \dfrac{1}{3}$

Common difference $d = \dfrac{5}{3} - \dfrac{1}{3} = \dfrac{4}{3}$

(iv) $0.6, 1.7, 2.8, 3.9, \ldots$

First term $a = 0.6$

Common difference $d = 1.7 - 0.6 = 1.1$

(i) $2, 4, 8, 16, \ldots$

$a_2 - a_1 = 4 - 2 = 2$
$a_3 - a_2 = 8 - 4 = 4$

Since $2 \neq 4$, this is NOT an AP.

(ii) $2, \dfrac{5}{2}, 3, \dfrac{7}{2}, \ldots$

$a_2 - a_1 = \dfrac{5}{2} - 2 = \dfrac{1}{2}$
$a_3 - a_2 = 3 - \dfrac{5}{2} = \dfrac{1}{2}$
$a_4 - a_3 = \dfrac{7}{2} - 3 = \dfrac{1}{2}$

This is an AP with $d = \dfrac{1}{2}$.

Next three terms:
$a_5 = \dfrac{7}{2} + \dfrac{1}{2} = 4$
$a_6 = 4 + \dfrac{1}{2} = \dfrac{9}{2}$
$a_7 = \dfrac{9}{2} + \dfrac{1}{2} = 5$

Three more terms: $4,\ \dfrac{9}{2},\ 5$

(iii) $-1.2, -3.2, -5.2, -7.2, \ldots$

$a_2 - a_1 = -3.2 - (-1.2) = -2$
$a_3 - a_2 = -5.2 - (-3.2) = -2$
$a_4 - a_3 = -7.2 - (-5.2) = -2$

This is an AP with $d = -2$.

Next three terms:
$a_5 = -7.2 + (-2) = -9.2$
$a_6 = -11.2$
$a_7 = -13.2$

Three more terms: $-9.2,\ -11.2,\ -13.2$

(iv) $-10, -6, -2, 2, \ldots$

$a_2 - a_1 = -6 - (-10) = 4$
$a_3 - a_2 = -2 - (-6) = 4$
$a_4 - a_3 = 2 - (-2) = 4$

This is an AP with $d = 4$.

Next three terms:
$a_5 = 2 + 4 = 6$
$a_6 = 10$
$a_7 = 14$

Three more terms: $6,\ 10,\ 14$

(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots$

$a_2 - a_1 = (3+\sqrt{2}) - 3 = \sqrt{2}$
$a_3 - a_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2}$
$a_4 - a_3 = \sqrt{2}$

This is an AP with $d = \sqrt{2}$.

Next three terms:
$a_5 = 3+4\sqrt{2}$
$a_6 = 3+5\sqrt{2}$
$a_7 = 3+6\sqrt{2}$

Three more terms: $3+4\sqrt{2},\ 3+5\sqrt{2},\ 3+6\sqrt{2}$

(vi) $0.2, 0.22, 0.222, 0.2222, \ldots$

$a_2 - a_1 = 0.22 - 0.2 = 0.02$
$a_3 - a_2 = 0.222 - 0.22 = 0.002$

Since $0.02 \neq 0.002$, this is NOT an AP.

(vii) $0, -4, -8, -12, \ldots$

$a_2 - a_1 = -4 - 0 = -4$
$a_3 - a_2 = -8 - (-4) = -4$
$a_4 - a_3 = -12 - (-8) = -4$

This is an AP with $d = -4$.

Next three terms:
$a_5 = -12 + (-4) = -16$
$a_6 = -20$
$a_7 = -24$

Three more terms: $-16,\ -20,\ -24$

(viii) $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, \ldots$

$a_2 - a_1 = 0$, $a_3 - a_2 = 0$, $a_4 - a_3 = 0$

This is an AP with $d = 0$.

Next three terms: $-\dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{1}{2}$

(ix) $1, 3, 9, 27, \ldots$

$a_2 - a_1 = 3 - 1 = 2$
$a_3 - a_2 = 9 - 3 = 6$

Since $2 \neq 6$, this is NOT an AP.

(x) $a, 2a, 3a, 4a, \ldots$

$a_2 - a_1 = 2a - a = a$
$a_3 - a_2 = 3a - 2a = a$
$a_4 - a_3 = 4a - 3a = a$

This is an AP with $d = a$.

Next three terms:
$a_5 = 5a,\quad a_6 = 6a,\quad a_7 = 7a$

Three more terms: $5a,\ 6a,\ 7a$

(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$

Simplify: $\sqrt{2},\ 2\sqrt{2},\ 3\sqrt{2},\ 4\sqrt{2}, \ldots$

$a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$
$a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$
$a_4 - a_3 = \sqrt{2}$

This is an AP with $d = \sqrt{2}$.

Next three terms:
$a_5 = 5\sqrt{2} = \sqrt{50}$
$a_6 = 6\sqrt{2} = \sqrt{72}$
$a_7 = 7\sqrt{2} = \sqrt{98}$

Three more terms: $\sqrt{50},\ \sqrt{72},\ \sqrt{98}$

(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$

$a_2 - a_1 = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2}-1)$
$a_3 - a_2 = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6}$

Since $\sqrt{3}(\sqrt{2}-1) \approx 0.732$ and $3 - \sqrt{6} \approx 0.551$, these are not equal.

This is NOT an AP.

(xiv) $1^2, 5^2, 7^2, 73, \ldots$ i.e., $1, 25, 49, 73, \ldots$

$a_2 - a_1 = 25 - 1 = 24$
$a_3 - a_2 = 49 - 25 = 24$
$a_4 - a_3 = 73 - 49 = 24$

This is an AP with $d = 24$.

Next three terms:
$a_5 = 73 + 24 = 97$
$a_6 = 121$
$a_7 = 145$

Three more terms: $97,\ 121,\ 145$

Using the formula $a_n = a + (n-1)d$:

(i) $a=7$, $d=3$, $n=8$

$$a_8 = 7 + (8-1)\times 3 = 7 + 21 = 28$$

$\boxed{a_n = 28}$

(ii) $a=-18$, $n=10$, $a_{10}=0$

$$0 = -18 + (10-1)d$$
$$9d = 18$$
$$d = 2$$

$\boxed{d = 2}$

(iii) $d=-3$, $n=18$, $a_{18}=-5$

$$-5 = a + (18-1)\times(-3)$$
$$-5 = a - 51$$
$$a = 46$$

$\boxed{a = 46}$

(iv) $a=-18.9$, $d=2.5$, $a_n=3.6$

$$3.6 = -18.9 + (n-1)\times 2.5$$
$$(n-1)\times 2.5 = 3.6 + 18.9 = 22.5$$
$$n-1 = 9$$
$$n = 10$$

$\boxed{n = 10}$

(v) $a=3.5$, $d=0$, $n=105$

$$a_{105} = 3.5 + (105-1)\times 0 = 3.5$$

$\boxed{a_n = 3.5}$

(i) AP: 10, 7, 4, …

$a = 10$, $d = 7 - 10 = -3$

$$a_{30} = a + (30-1)d = 10 + 29\times(-3) = 10 - 87 = -77$$

Correct option: (C) −77

(ii) AP: $-3, -\dfrac{1}{2}, 2, \ldots$

$a = -3$, $d = -\dfrac{1}{2} - (-3) = -\dfrac{1}{2} + 3 = \dfrac{5}{2}$

$$a_{11} = -3 + (11-1)\times\frac{5}{2} = -3 + 10\times\frac{5}{2} = -3 + 25 = 22$$

Correct option: (B) 22

(i) 2, □, 26

Here $a_1 = 2$, $a_3 = 26$.

$a_3 = a + 2d \Rightarrow 26 = 2 + 2d \Rightarrow d = 12$

$a_2 = 2 + 12 = 14$

Missing term: 14

(ii) □, 13, □, 3

$a_2 = 13$, $a_4 = 3$

$a_4 = a_2 + 2d \Rightarrow 3 = 13 + 2d \Rightarrow d = -5$

$a_1 = 13 - (-5) = 18$

$a_3 = 13 + (-5) = 8$

Missing terms: 18, 8

(iii) 5, □, □, $9\dfrac{1}{2}$

$a_1 = 5$, $a_4 = \dfrac{19}{2}$

$a_4 = a + 3d \Rightarrow \dfrac{19}{2} = 5 + 3d \Rightarrow 3d = \dfrac{9}{2} \Rightarrow d = \dfrac{3}{2}$

$a_2 = 5 + \dfrac{3}{2} = \dfrac{13}{2} = 6.5$

$a_3 = 5 + 2\times\dfrac{3}{2} = 5 + 3 = 8$

Missing terms: $6\dfrac{1}{2}$, $8$

(iv) $-4$, □, □, □, □, $6$

$a_1 = -4$, $a_6 = 6$

$a_6 = a + 5d \Rightarrow 6 = -4 + 5d \Rightarrow d = 2$

$a_2 = -4+2 = -2$
$a_3 = -2+2 = 0$
$a_4 = 0+2 = 2$
$a_5 = 2+2 = 4$

Missing terms: −2, 0, 2, 4

(v) □, 38, □, □, □, $-22$

$a_2 = 38$, $a_6 = -22$

$a_6 = a_2 + 4d \Rightarrow -22 = 38 + 4d \Rightarrow 4d = -60 \Rightarrow d = -15$

$a_1 = 38 - (-15) = 53$
$a_3 = 38 + (-15) = 23$
$a_4 = 23 + (-15) = 8$
$a_5 = 8 + (-15) = -7$

Missing terms: 53, 23, 8, −7

Given AP: 3, 8, 13, 18, …

$a = 3$, $d = 8 - 3 = 5$

Let $a_n = 78$.

Using $a_n = a + (n-1)d$:

$$78 = 3 + (n-1)\times 5$$
$$75 = (n-1)\times 5$$
$$n - 1 = 15$$
$$n = 16$$

78 is the 16th term of the AP.

(i) 7, 13, 19, …, 205

$a = 7$, $d = 13 - 7 = 6$, $a_n = 205$

$$205 = 7 + (n-1)\times 6$$
$$198 = (n-1)\times 6$$
$$n - 1 = 33$$
$$n = 34$$

There are 34 terms.

(ii) $18, 15\dfrac{1}{2}, 13, \ldots, -47$

$a = 18$, $d = 15.5 - 18 = -2.5$, $a_n = -47$

$$-47 = 18 + (n-1)\times(-2.5)$$
$$-65 = (n-1)\times(-2.5)$$
$$n - 1 = 26$$
$$n = 27$$

There are 27 terms.

Given AP: 11, 8, 5, 2, …

$a = 11$, $d = 8 - 11 = -3$

Assume $a_n = -150$.

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

Since $n$ is not a natural number, −150 is NOT a term of this AP.

Given: $a_{11} = 38$ and $a_{16} = 73$.

Using $a_n = a + (n-1)d$:

$$a + 10d = 38 \quad \cdots (1)$$
$$a + 15d = 73 \quad \cdots (2)$$

Subtracting (1) from (2):
$$5d = 35 \Rightarrow d = 7$$

From (1): $a = 38 - 70 = -32$

$$a_{31} = a + 30d = -32 + 30\times 7 = -32 + 210 = 178$$

The 31st term is 178.

Given: $n = 50$, $a_3 = 12$, $a_{50} = 106$.

$$a + 2d = 12 \quad \cdots (1)$$
$$a + 49d = 106 \quad \cdots (2)$$

Subtracting (1) from (2):
$$47d = 94 \Rightarrow d = 2$$

From (1): $a = 12 - 4 = 8$

$$a_{29} = a + 28d = 8 + 28\times 2 = 8 + 56 = 64$$

The 29th term is 64.

Given: $a_3 = 4$ and $a_9 = -8$.

$$a + 2d = 4 \quad \cdots (1)$$
$$a + 8d = -8 \quad \cdots (2)$$

Subtracting (1) from (2):
$$6d = -12 \Rightarrow d = -2$$

From (1): $a = 4 - 2(-2) = 4 + 4 = 8$

Let $a_n = 0$:
$$0 = 8 + (n-1)\times(-2)$$
$$(n-1)\times 2 = 8$$
$$n - 1 = 4 \Rightarrow n = 5$$

The 5th term of this AP is zero.

Given: $a_{17} - a_{10} = 7$

$$[a + 16d] - [a + 9d] = 7$$
$$7d = 7$$
$$d = 1$$

The common difference is 1.

Given AP: 3, 15, 27, 39, …

$a = 3$, $d = 15 - 3 = 12$

First, find the 54th term:
$$a_{54} = 3 + 53\times 12 = 3 + 636 = 639$$

Let the required term be $a_n$:
$$a_n = a_{54} + 132 = 639 + 132 = 771$$

$$771 = 3 + (n-1)\times 12$$
$$768 = (n-1)\times 12$$
$$n - 1 = 64 \Rightarrow n = 65$$

The 65th term is 132 more than the 54th term.

Let the two APs have first terms $a$ and $b$ respectively, with the same common difference $d$.

100th term of 1st AP: $a + 99d$
100th term of 2nd AP: $b + 99d$

Given: $(a + 99d) - (b + 99d) = 100$
$$\Rightarrow a - b = 100$$

1000th term of 1st AP: $a + 999d$
1000th term of 2nd AP: $b + 999d$

Difference $= (a + 999d) - (b + 999d) = a - b = 100$

The difference between their 1000th terms is also 100.

The smallest three-digit number divisible by 7:
$100 \div 7 = 14$ remainder $2$, so smallest = $100 + (7-2) = 105$

The largest three-digit number divisible by 7:
$999 \div 7 = 142$ remainder $5$, so largest = $999 - 5 = 994$

AP: 105, 112, 119, …, 994 with $a = 105$, $d = 7$, $a_n = 994$

$$994 = 105 + (n-1)\times 7$$
$$889 = (n-1)\times 7$$
$$n - 1 = 127 \Rightarrow n = 128$$

There are 128 three-digit numbers divisible by 7.

The smallest multiple of 4 greater than 10 is 12.
The largest multiple of 4 less than 250 is 248.

AP: 12, 16, 20, …, 248 with $a = 12$, $d = 4$, $a_n = 248$

$$248 = 12 + (n-1)\times 4$$
$$236 = (n-1)\times 4$$
$$n - 1 = 59 \Rightarrow n = 60$$

There are 60 multiples of 4 between 10 and 250.

AP 1: 63, 65, 67, … → $a_1 = 63$, $d_1 = 2$

$n$th term: $63 + (n-1)\times 2 = 61 + 2n$

AP 2: 3, 10, 17, … → $a_2 = 3$, $d_2 = 7$

$n$th term: $3 + (n-1)\times 7 = -4 + 7n$

Setting them equal:
$$61 + 2n = -4 + 7n$$
$$65 = 5n$$
$$n = 13$$

The 13th terms of the two APs are equal.

Given: $a_3 = 16$ and $a_7 - a_5 = 12$.

From $a_7 - a_5 = 12$:
$$[a + 6d] - [a + 4d] = 12$$
$$2d = 12 \Rightarrow d = 6$$

From $a_3 = 16$:
$$a + 2d = 16$$
$$a + 12 = 16 \Rightarrow a = 4$$

The AP is: $4, 10, 16, 22, 28, \ldots$

The AP is 4, 10, 16, 22, 28, …

Given AP: 3, 8, 13, …, 253

$a = 3$, $d = 5$, last term $l = 253$

The 20th term from the last term is the same as the 20th term of the AP written in reverse, which has first term 253 and common difference $-5$.

$$a_{20}\text{ (from last)} = 253 + (20-1)\times(-5) = 253 - 95 = 158$$

The 20th term from the last is 158.

Given: $a_4 + a_8 = 24$ and $a_6 + a_{10} = 44$.

$$a_4 + a_8 = (a+3d) + (a+7d) = 2a + 10d = 24$$
$$\Rightarrow a + 5d = 12 \quad \cdots (1)$$

$$a_6 + a_{10} = (a+5d) + (a+9d) = 2a + 14d = 44$$
$$\Rightarrow a + 7d = 22 \quad \cdots (2)$$

Subtracting (1) from (2):
$$2d = 10 \Rightarrow d = 5$$

From (1): $a = 12 - 25 = -13$

First three terms:
$a_1 = -13$
$a_2 = -13 + 5 = -8$
$a_3 = -8 + 5 = -3$

The first three terms are −13, −8, −3.

Given: $a = 5000$, $d = 200$, $a_n = 7000$.

$$7000 = 5000 + (n-1)\times 200$$
$$2000 = (n-1)\times 200$$
$$n - 1 = 10 \Rightarrow n = 11$$

The 11th year from 1995 is $1995 + 10 = 2005$.

Subba Rao's income reached ₹ 7000 in the year 2005.

Given: $a = 5$, $d = 1.75$, $a_n = 20.75$.

$$20.75 = 5 + (n-1)\times 1.75$$
$$15.75 = (n-1)\times 1.75$$
$$n - 1 = \frac{15.75}{1.75} = 9$$
$$n = 10$$

In the 10th week, her savings become ₹ 20.75.

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$:

(i) $a = 2$, $d = 5$, $n = 10$

$$S_{10} = \frac{10}{2}[2\times2 + 9\times5] = 5[4 + 45] = 5\times49 = 245$$

Sum = 245

(ii) $a = -37$, $d = 4$, $n = 12$

$$S_{12} = \frac{12}{2}[2\times(-37) + 11\times4] = 6[-74 + 44] = 6\times(-30) = -180$$

Sum = −180

(iii) $a = 0.6$, $d = 1.1$, $n = 100$

$$S_{100} = \frac{100}{2}[2\times0.6 + 99\times1.1] = 50[1.2 + 108.9] = 50\times110.1 = 5505$$

Sum = 5505

(iv) $a = \dfrac{1}{15}$, $d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5-4}{60} = \dfrac{1}{60}$, $n = 11$

$$S_{11} = \frac{11}{2}\left[2\times\frac{1}{15} + 10\times\frac{1}{60}\right] = \frac{11}{2}\left[\frac{2}{15} + \frac{1}{6}\right]$$

$$= \frac{11}{2}\left[\frac{4}{30} + \frac{5}{30}\right] = \frac{11}{2}\times\frac{9}{30} = \frac{11}{2}\times\frac{3}{10} = \frac{33}{20}$$

Sum = $\dfrac{33}{20}$