Find the sum of the products of the corresponding terms of finite geometrical progressions
2, 4, 8, 16, 32 and 128, 32, 8, 2, `1/2`.

Explanation

Raking the products of the corresponding terms of two given GPs, we get a new progression
`(2xx128), (4xx32), (8xx8), (16xx2), (32xx1/2)`
i.e., `256, 128, 64, 32, 16`, which is clealy a GP in which `a=256` and `r=16/32=1/2 lt 1`.
`:.` the required sum `=(a(1-r^(5)))/((1-r))=(256xx{1-(1/2)^(5)})/((1-1/2))`
`=(256xx(1-1/2^(5)))/((1/2))=256xx2xx(1-1/32)`
`=(256xx2xx31/32)=496`.