NCERT Solutions for Class 9 Maths Chapter 8 — End-of-Chapter Exercises#

The 11th term is 38, so:

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

The 16th term is 73, so:

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

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 15d &=& 73 \\ a + 10d &=& 38 \\ \hline 5d &=& 35 \end{array}$$

$$\Rightarrow d = 7$$

Substituting $d = 7$ in equation (1):

$$a + 10(7) = 38$$

$$\Rightarrow a = -32$$

So:

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

$$t_{31} = -32 + 210$$

$$t_{31} = 178$$

∴ The 31st term is 178.

The 7th term exceeds the 5th term by 12, so:

$$(a + 6d) - (a + 4d) = 12$$

$$\Rightarrow 2d = 12$$

$$\Rightarrow d = 6$$

The third term is 16, so:

$$a + 2d = 16$$

$$a + 12 = 16$$

$$\Rightarrow a = 4$$

∴ The AP is 4, 10, 16, 22, 28, …

The smallest three-digit multiple of 7 is $7 \times 15 = 105$, and the largest is $7 \times 142 = 994$. They form an AP with $a = 105$, $d = 7$ and last term 994:

$$105 + (n - 1) \times 7 = 994$$

$$\Rightarrow (n - 1) \times 7 = 889$$

$$\Rightarrow n - 1 = 127$$

$$\Rightarrow n = 128$$

∴ There are 128 three-digit numbers divisible by 7.

The smallest multiple of 4 above 10 is 12, and the largest below 250 is 248. They form an AP with $a = 12$, $d = 4$ and last term 248:

$$12 + (n - 1) \times 4 = 248$$

$$\Rightarrow (n - 1) \times 4 = 236$$

$$\Rightarrow n - 1 = 59$$

$$\Rightarrow n = 60$$

∴ There are 60 multiples of 4 between 10 and 250.

Let the first term be $a$ and the ratio $r$ ($a \neq 0$).

The fifth term is 4 times the third term, so:

$$ar^4 = 4ar^2$$

$$\Rightarrow r^2 = 4 \quad (\text{dividing by } ar^2,\ r \neq 0)$$

$$\Rightarrow r = 2 \text{ or } r = -2$$

The sum of the first two terms is $-4$, so:

$$a + ar = -4$$

$$\Rightarrow a(1 + r) = -4$$

If $r = 2$: $3a = -4$, so $a = -\dfrac{4}{3}$. The GP is

$$-\dfrac{4}{3}, \ -\dfrac{8}{3}, \ -\dfrac{16}{3}, \ -\dfrac{32}{3}, \ \ldots$$

If $r = -2$: $-a = -4$, so $a = 4$. The GP is

$$4, \ -8, \ 16, \ -32, \ 64, \ \ldots$$

∴ Either GP works: 4, −8, 16, −32, … or $-\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}, \ldots$

(Check for the second: $4 + (-8) = -4$ ✓, and $t_5 = 64 = 4 \times 16 = 4t_3$ ✓.)

Suppose 100 is the sum of $m$ consecutive natural numbers starting from $k$. This is an AP with $d = 1$, so:

$$\dfrac{m}{2}\bigl(2k + (m - 1)\bigr) = 100$$

$$\Rightarrow m(2k + m - 1) = 200$$

$m$ and $2k + m - 1$ multiply to 200, and one is odd while the other is even (their difference $2k - 1$ is odd). Also $2k + m - 1 > m$ (since $k \geq 1$). The factor pairs of 200 with one odd factor are $1 \times 200$, $5 \times 40$ and $8 \times 25$:

$m$	$2k + m - 1$	$k$	Sum
5	40	18	$18 + 19 + 20 + 21 + 22$
8	25	9	$9 + 10 + 11 + \cdots + 16$

($m = 1$ gives just “100” on its own, which is not a sum of several numbers.)

∴ There are two ways:

$$100 = 18 + 19 + 20 + 21 + 22$$

$$100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$$

The number at the end of each hour forms a GP: start with 30 and multiply by 2 every hour. At the end of hour $n$:

$$B_n = 30 \times 2^n$$$$B_2 = 30 \times 2^2 = 120$$

$$B_4 = 30 \times 2^4 = 480$$

∴ There are 120 bacteria after 2 hours, 480 after 4 hours, and $30 \times 2^n$ after $n$ hours.

The sum of the 4th and 8th terms is 24, so:

$$(a + 3d) + (a + 7d) = 24$$

$$\Rightarrow 2a + 10d = 24 \quad \cdots (1)$$

The sum of the 6th and 10th terms is 44, so:

$$(a + 5d) + (a + 9d) = 44$$

$$\Rightarrow 2a + 14d = 44 \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 2a + 14d &=& 44 \\ 2a + 10d &=& 24 \\ \hline 4d &=& 20 \end{array}$$

$$\Rightarrow d = 5$$

Substituting $d = 5$ in equation (1):

$$2a + 50 = 24$$

$$\Rightarrow a = -13$$

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

We need $\dfrac{n(n + 1)}{2} > 1000$, that is, $n(n + 1) > 2000$. Since $\sqrt{2000} \approx 44.7$, we try values near 44:

$$n = 44: \quad \dfrac{44 \times 45}{2} = 990 \quad (\text{not greater than } 1000)$$

$$n = 45: \quad \dfrac{45 \times 46}{2} = 1035 \quad (\text{greater than } 1000)$$

∴ The smallest value is $n = 45$.

Here $a = 2$ and $r = 4$:

$$t_n = 2 \times 4^{\,n-1}$$

We set it equal to 131072:

$$2 \times 4^{\,n-1} = 131072$$

$$\Rightarrow 4^{\,n-1} = 65536$$

$$\Rightarrow 4^{\,n-1} = 4^8$$

$$\Rightarrow n = 9$$

∴ 131072 is the 9th term.

• Explicit formula: $t_n = 2 \times 4^{\,n-1}$
• Recursive formula: $t_1 = 2$, $t_{n+1} = 4t_n$ for $n \geq 1$

Let the three terms be $\dfrac{a}{r}$, $a$, $ar$.

Their product is $-1$, so:

$$\dfrac{a}{r} \times a \times ar = -1$$

$$\Rightarrow a^3 = -1$$

$$\Rightarrow a = -1$$

Their sum is $\dfrac{13}{12}$, so:

$$-\dfrac{1}{r} - 1 - r = \dfrac{13}{12}$$

Multiplying by $-12r$:

$$12 + 12r + 12r^2 = -13r$$

$$\Rightarrow 12r^2 + 25r + 12 = 0$$

We need two numbers with sum 25 and product $12 \times 12 = 144$: they are 16 and 9. We factor $12r^2 + 25r + 12$:

$$= 12r^2 + 16r + 9r + 12$$

$$= 4r(3r + 4) + 3(3r + 4)$$

$$= (3r + 4)(4r + 3)$$$$\Rightarrow r = -\dfrac{4}{3} \quad \text{or} \quad r = -\dfrac{3}{4}$$

If $r = -\dfrac{4}{3}$: the terms are $\dfrac{3}{4}, -1, \dfrac{4}{3}$.

If $r = -\dfrac{3}{4}$: the terms are $\dfrac{4}{3}, -1, \dfrac{3}{4}$.

∴ The common ratio is $-\dfrac{4}{3}$ or $-\dfrac{3}{4}$, and the terms are $\dfrac{3}{4}, -1, \dfrac{4}{3}$ (in either order). Check: $\dfrac{3}{4} - 1 + \dfrac{4}{3} = \dfrac{9 - 12 + 16}{12} = \dfrac{13}{12}$ ✓

Let the GP have first term $a$ and ratio $r$. Then:

$$x = ar^3, \quad y = ar^9, \quad z = ar^{15}$$$$\dfrac{y}{x} = \dfrac{ar^9}{ar^3} = r^6$$

$$\dfrac{z}{y} = \dfrac{ar^{15}}{ar^9} = r^6$$$$\Rightarrow \dfrac{y}{x} = \dfrac{z}{y}$$

∴ $x$, $y$, $z$ have a common ratio $r^6$, so they are in GP. (Equivalently, $y^2 = a^2r^{18} = xz$.)

Let the terms be $a$, $ar$, $ar^2$.

The sum is 26, so:

$$a(1 + r + r^2) = 26 \quad \cdots (1)$$

The sum of the squares is 364, so:

$$a^2(1 + r^2 + r^4) = 364 \quad \cdots (2)$$

We use the identity $1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)$, which we can check by expanding. Dividing equation (2) by the square of equation (1):

$$\dfrac{a^2(1 + r + r^2)(1 - r + r^2)}{a^2(1 + r + r^2)^2} = \dfrac{364}{676}$$

$$\Rightarrow \dfrac{1 - r + r^2}{1 + r + r^2} = \dfrac{7}{13}$$

$$\Rightarrow 13 - 13r + 13r^2 = 7 + 7r + 7r^2$$

$$\Rightarrow 6r^2 - 20r + 6 = 0$$

$$\Rightarrow 3r^2 - 10r + 3 = 0$$

$$\Rightarrow (3r - 1)(r - 3) = 0$$

So $r = 3$ or $r = \dfrac{1}{3}$.

If $r = 3$: from equation (1), $a(1 + 3 + 9) = 26$, so $a = 2$. The terms are 2, 6, 18.

If $r = \dfrac{1}{3}$: $a\left(1 + \dfrac{1}{3} + \dfrac{1}{9}\right) = 26$, so $a = 18$. The terms are 18, 6, 2.

∴ The terms are 2, 6, 18 (or the same in reverse order, 18, 6, 2). Check: $2 + 6 + 18 = 26$ ✓ and $4 + 36 + 324 = 364$ ✓

$$P_3 = P_1 + P_2 + 1 = 1 + 2 + 1 = 4$$

$$P_4 = 1 + 2 + 4 + 1 = 8$$

$$P_5 = 1 + 2 + 4 + 8 + 1 = 16$$

Continuing: $P_6 = 32$, $P_7 = 64$, $P_8 = 128$.

∴ $P_1, \ldots, P_8$ are 1, 2, 4, 8, 16, 32, 64, 128.

Simpler recursive formula: for $n > 2$,

$$P_{n+1} = (P_1 + \cdots + P_{n-1} + 1) + P_n$$

$$P_{n+1} = P_n + P_n$$

$$P_{n+1} = 2P_n$$

So $P_1 = 1$, $P_{n+1} = 2P_n$ (this also works for $P_2 = 2P_1$).

Explicit formula: $P_n = 2^{\,n-1}$.

$$W_3 = W_1 + 2 = 1 + 2 = 3$$

$$W_4 = W_1 + W_2 + 2 = 1 + 2 + 2 = 5$$

$$W_5 = 1 + 2 + 3 + 2 = 8$$

$$W_6 = 1 + 2 + 3 + 5 + 2 = 13$$

$$W_7 = 1 + 2 + 3 + 5 + 8 + 2 = 21$$

$$W_8 = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34$$

∴ $W_1, \ldots, W_8$ are 1, 2, 3, 5, 8, 13, 21, 34.

This is the Virahānka–Fibonacci sequence: each term is the sum of the two before it. Indeed, for $n > 3$:

$$W_n - W_{n-1} = W_{n-2}$$

$$\Rightarrow W_n = W_{n-1} + W_{n-2}$$