Let a , a r , a r 2 , … be an infinite G.P. If ? n = 0 ? a r n = 57 and ? n = 0 ? a 3 r 3 n = 9747 , then a + 18 r is equal to [2024]

Question

Let $a, a r, a r^{2}, \dots$ be an infinite G.P. If $\sum_{n = 0}^{\infty} a r^{n} = 57$ and $\sum_{n = 0}^{\infty} a^{3} r^{3 n} = 9747,$ then $a + 18 r$ is equal to [2024]

Smart Achievers · Accepted Answer

(3)

Given, $\sum_{n = 0}^{\infty} a r^{n} = 57$

$\Rightarrow \frac{a}{1 - r} = 57$ ...(i)

[Sum of infinite G.P.]

Also, $\sum_{n = 0}^{\infty} a^{3} r^{3 n} = 9747$

i.e., $a^{3} + a^{3} r^{3} + \dots \infty = 9747$

$\Rightarrow \frac{a^{3}}{1 - r^{3}} = 9747 \Rightarrow \frac{(57 {(1 - r))}^{3}}{1 - r^{3}} = 9747$

$\frac{(1 - r) (1 + r + r^{2})}{{(1 - r)}^{3}} = 19 \Rightarrow 19 {(1 - r)}^{2} = 1 + r + r^{2}$

$\Rightarrow 19 + 19 r^{2} - 38 r = 1 + r + r^{2}$

$\Rightarrow 18 r^{2} - 39 r + 18 = 0$

$\Rightarrow r = \frac{2}{3}, \frac{3}{2} \Rightarrow r = \frac{2}{3}$ $[∵ | r | < 1]$

$∴$ $a = 57 (1 - \frac{2}{3}) \Rightarrow a = 19$

So, $a + 18 r = 19 + 12 = 31$

Solution

Q.
Let $a, a r, a r^{2}, \dots$ be an infinite G.P. If $\sum_{n = 0}^{\infty} a r^{n} = 57$ and $\sum_{n = 0}^{\infty} a^{3} r^{3 n} = 9747,$ then $a + 18 r$ is equal to [2024]

1 38

2 46

3 31

4 27

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