The `1^(st)` , `2^(nd)` and `3^(rd)` terms of an arithmetic series are `a`, `b` and `a^(2)` where `'a'` is negative. The `1^(st)`, `2^(nd)` and `3^(rd)` terms of a geometric series are `a`, `a^(2)` and `b` respectively.
The sum of the `40` terms of the arithmetic series is

Options

A. `(545)/(2)`
B. `220`
C. `250`
D. `(575)/(2)`

Answer

A

Explanation

`(a)` `a,b,a^(2)`,……is an `A.P.`
`:. 2b=a^(2)+a`…..`(i)`
`a,a^(2),b`,………is a `G.P.`
`:.a^(3)=b`…….`(ii)`
`:.a^(2)+a=2a^(3)`
`:.a=0`, `a=1` or `a=-1//2`
`:.a=-1//2` (as `a lt 0`)
`:.b=-(1)/(8)`
In `G.P.` putting `a=-(1)/(2)` and `b=-(1)/(8)`
`:. G.P.` is `-(1)/(2),(1)/(4),-(1)/(8)`...
Sum of infinite `G.P.=(a)/(1-r)=(-(1)/(2))/(1-(((1)/(4))/((-1)/(2))))=(-1)/(3)`
Sum of `40` terms of `A.P.=(n)/(2)[2a+(n-1)d]`
`=(40)/(2)[2((-1)/(2))+(40-1)xx(-(1)/(8)+(1)/(2))]`
`=20[-1+39xx(3)/(8)]`
`=(545)/(2)`