Question

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

Answer 1

Required sum = $2\times128+4$ $\times 32 +8\times8+16\times2+32\times \frac{1}{2}$

= $64[4+2+1+\frac{1}{2}+\frac{1}{22}]$

Here, 4, 2, 1, $\frac{1}{2},\frac{1}{2^2}$ is a G.P.

First term, a = 4

Common ratio, r = $\frac{1}{2}$

It is known that, $s_n$ = $\frac{a(1-r^n)}{1-r}$

∴ $S_5$= $\frac {4[1-(\frac 12)^5]}{1- \frac 12}$ = $\frac{4[1-\frac{1}{32}]}{\frac{1}{2}}$ = $8(\frac{32-1}{32})$= $\frac{31}{4}$

∴Required sum = $64(\frac{31}{4})$= (16)(31) = 496