If each term of a geometric progression a 1 , a 2 , a 3 , … with a 1 = 1 8 and a 2 ? a 1 , is the arithmetic mean of the next two terms and S n = a 1 + a 2 + … + a n , then S 20 - S 18 is equal to [2024]

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Solution

Q.
If each term of a geometric progression $a_{1}, a_{2}, a_{3}, \dots$ with $a_{1} = \frac{1}{8}$ and $a_{2} \neq a_{1},$ is the arithmetic mean of the next two terms and $S_{n} = a_{1} + a_{2} + \dots + a_{n},$ then $S_{20} - S_{18}$ is equal to [2024]

1 $- 2^{15}$

2 $2^{18}$

3 $2^{15}$

4 $- 2^{18}$

Ans.
(1)

Let $r$ be the common ratio.

Now, $a_{n} = A.M. of a_{n + 1} and a_{n + 2} = \frac{a_{n + 1} + a_{n + 2}}{2}$

$\Rightarrow a_{1} \cdot r^{n - 1} = \frac{1}{2} [a_{1} \cdot r^{n} + a_{1} \cdot r^{n + 1}] \Rightarrow 2 r^{n - 1} = r^{n} + r^{n + 1}$

$\Rightarrow 2 = r + r^{2} \Rightarrow r^{2} + r - 2 = 0 \Rightarrow (r + 2) (r - 1) = 0$

$\Rightarrow r = - 2 (∵ r \neq 1 as a_{1} \neq a_{2})$

Now, $S_{20} - S_{18} = a_{19} + a_{20} (∵ S_{n} = a_{1} + a_{2} + \dots + a_{n})$

$= \frac{1}{8} \cdot r^{18} + \frac{1}{8} \cdot r^{19} = \frac{1}{8} {(- 2)}^{18} [1 - 2] = \frac{- 2^{18}}{2^{3}} = - 2^{15}$

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