1.

The sum of n terms of a progression is `(2^(n)-1)`. Show that it is a GP and find its common ratio.

Answer» Correct Answer - `r=2`
`S_(n)=(2^(n)-1)` and `S_(n-1)=[2^((n-1))-1]`.
`:. T_(n)=(S_(n)-S_(n-1))=[2^(n)-2^((n-1))]rArr T_(1)=1, T_(2)=2, T_(3)=4, T_(4)=8, ...`
Clearly, 1, 2, 4, 8, ... is a GP with common ratio =2


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