Question

Find the sum of the products of the corresponding terms of finite geometrical progressions

2, 4, 8, 16, 32 and 128, 32, 8, 2, $\frac{1}{2}$

Answer 1

Taking the products of the corresponding terms of two given GPs, we get a new progression

$(2\times 128),(4\times 32),(8\times 8),(16\times 2),(32\times \frac{1}{2})$

i.e., 256, 128, 64, 32, 16, which is clearly a GP in which a= 256 and

$r=\frac{16}{32}=\frac{1}{2}<1$

$\therefore$ the required sum $=\frac{a(1-r^5)}{(1-r)}=\frac{256\times \{1-{\left(\frac{1}{2}\right)}^5\}}{(1-\frac{1}{2})}$

$=\frac{256\times (1-\frac{1}{2^5})}{\left(\frac{1}{2}\right)}=256\times 2\times (1-\frac{1}{32})$

$=(256\times 2\times \frac{31}{32})=496$