In a `GP`, the ratio of the sum of the first eleven terms of the sum of the last even terms is `1//8` and the ratio of the sum of all the terms without the first nine to the sum of all terms without the last nine is `2`. Then the number of terms in the `GP` is

Options

A. `40`
B. `38`
C. `36`
D. `34`

Answer

B

Explanation

`(b)` Let `G.P.` be `a,ar,ar^(3),….,ar^(n-1)`
`S_(11)=(a(1-r^(11))/(1-r)`
`S_(11)("from last")=ar^(n-11)((1-r^(11)))/(1-r)`
`implies(S_(11))/(S'_(11))=(1)/(r^(n-11))`
Given `(1)/(r^(n-11))=(1)/(8)`
`implies r^(n-11)=8`
Also for `ar^(9)...ar^(n-1)`, `S=ar^(9)*((1-r^(n-9)))/(1-r)`
and `S'` (sum of all the terms without the last mine) `=(a*(1-r^(n-9)))/(1-r)`
Now `(S)/(S')=r^(9)=2`(given)
`impliesr^(n-11)=2^(3)=(r^(9))^(3)`
`impliesn-11=27`
`impliesn=38`