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(A) The required sum is the sum of the products of corresponding terms: Sum $= (2 	imes 128) + (4 	imes 32) + (8 	imes 8) + (16 	imes 2) + (32 	i

Vedclass · Accepted Answer

$496$

Vedclass · Answer

(A) The required sum is the sum of the products of corresponding terms: Sum $= (2 	imes 128) + (4 	imes 32) + (8 	imes 8) + (16 	imes 2) + (32 	imes \frac{1}{2})$ Sum $= 256 + 128 + 64 + 32 + 16$ This is a geometric series with first term $a = 256$,common ratio $r = \frac{128}{256} = \frac{1}{2}$,and number of terms $n = 5$. The sum of $n$ terms of a $G.P.$ is given by $S_n = \frac{a(1 - r^n)}{1 - r}$. $S_5 = \frac{256(1 - (\frac{1}{2})^5)}{1 - \frac{1}{2}} = \frac{256(1 - \frac{1}{32})}{\frac{1}{2}} = 512 	imes \frac{31}{32} = 16 	imes 31 = 496$.

Find the sum of the products of the corresponding terms of the sequences $2, 4, 8, 16, 32$ and $128, 32, 8, 2, \frac{1}{2}$.

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