Question

If sum of first four terms of a geometric series is `30` and sum of the series to infinity is `32`, find the difference between the sum of the series to infinity and the sum of its first eight terms?

Answer 1

The difference between the sum to infinity and the sum of the first eight terms is `1/8`.

Explanation:

A geometric series is one in which ratio of a term to its preceding term, generally described as `r`, is always constant.

It is mentioned that sum to infinity is `32`, which means series is converging and `r<1`. In a geometric series, if `a` is the first term, such "sum to infinity" is given by `a/(1-r)`. Hence, `a/(1-r)=32`.

Further, sum of first `n` terms is given by `a×(1-r^n)/(1-r)`. As first four terms add up to `30`, `a×(1-r^4)/(1-r)=30`, but as `a/(1-r)=32`, we have

`32×(1-r^4)=30` or `1-r^4=30/32=15/16` and `r^4=1-15/16=1/16` or `r=1/2`.

Hence, as `a/(1-r)=32`, `a=32×(1-1/2)=32×1/2=16` i.e. first term is `16` and series is `{16,8,4,2,...}`.

First `8` terms add upto `16×(1-1/2^8)/(1-1/2)` or

`16(1-1/256)/(1/2)`

= `16×2×(1-1/256)`

= `32-1/8=255/8 or 31 7/8`

Hence, the difference between the sum to infinity, which is `32` and the sum of the first eight terms is `1/8`.