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
A
`(545)/(2)`
B
`220`
C
`250`
D
`(575)/(2)`
Text Solution
Verified by Experts
The correct Answer is:
A
`(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)`
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CENGAGE-PROGRESSION AND SERIES-ARCHIVES (MATRIX MATCH TYPE )