If `S_n` denotes the sum of first `n` terms of an A.P. and `(S_(3n)-S_(n-1))/(S_(2n)-S_(2n-1))=31` , then the value of `n` is 21`` b. 15`` c.16`` d. 19

Options

A. 21
B. 15
C. 16
D. 19

Answer

B

Explanation

`S_(3n)=(3n)/2[2a+(3n-1)d]`
`S_(n-1)=(n-1)/2[2a+(n-2)d]`
`rArrS_(3n)-S_(n-1)=1/2[2a(3n-n+1)]+d/2[3n(3n-1)-(n-1)(n-2)]`
`=1/2[2a(2n+1)+d(8n^(2)-2)]`
`=a(2n+1)+d(4n^(2)-1)`
=(2n+1)[a+(2n-1)d]
`S_(2n)-S_(2n-1)=T_(2n)=a+(2n-1)d`
`rArr(S_(3n)-S_(n-1))/(S_(2n)-S_(2n-1))=(2n+1)`
Given,
`((S_(3n)-S_(n-1))/(S_(2n)-S_(2n-1)))31`
`rArrn=15`