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
A
`40`
B
`38`
C
`36`
D
`34`
Text Solution
Verified by Experts
The correct Answer is:
B
`(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`
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CENGAGE-PROGRESSION AND SERIES-ARCHIVES (MATRIX MATCH TYPE )
