Question

The first three terms of a geometric progression are 6+n,10+n, and 15+n.find (I)the value of n? (II)the common ratio? (III)the sum of the first 6 terms of the sequence?

Answer 1

`n=10`. common ratio is `5/4` and sum of the first `6` terms of the sequence is `180 9/64`

Explanation:

As `6+n,10+n` and `15+n` are first three terms of a geometric progression,

we have `(10+n)^2=(6+n)(15+n)`

or `100+20n+n^2=90+21n+n^2`

or `21n-20n=100-90` i.e. `n=10`

and first three terms of a geometric progression are `16,20` and `25`.

i.e. first term is `16` and common ratio is `20/16=5/4`

As such sum of first `6` terms is `16xx((5/4)^6-1)/(5/4-1)`

= `16xx(15625/4096-1)/(1/4)`

= `16xx4xx11529/4096`

= `11529/64`

= `180 9/64`