Find four numbers forming a geometric progression.in which the third term is greater than first term by 9, and the second term 'is greater than the `4^ (th) ` by 18 .

Explanation

Four numbers are `a, ar, ar^2 ar^3`

`a r^2-a=9 rarr a(r^2-1)=9`

`a r-a r^3=18 rarr ar(1-r^2)=18 rarr` ar `(r^2-1)=-18`

`rarr r xx 9=-18 rarr r=-18/9=-2`


Substitute this value in `(1), a(4-1)=9 , therefore a=3`

`therefore` The required sequence is `3,-6,12,-24 . . . .`