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 a. 21`` b. 15`` c.16`` d. 19

Options

A. 2n-1
B. 2n+1
C. 4n+1
D. 2n+3

Answer

B

Explanation

We have,
`(S_(3n)-S_(n-1))/(S_(2n)-S_(n-1))=((3n)/(2){2a+(3n-1)d}-(n-1)/(2){2a+(n-2)d})/((2n)/(2){2a+(2n-1)d}-(2n-1)/(2){2a+(2n-2)d})`
`=(2a(3n-n+1)+{3n(3n-1)-(n-1)(n-2)}d)/(2a(2n-2n+1)+{2n(2n-1)-(2n-1)(2n-2)}d)`
`=(2a(2n+1)+{9n^(2)-3n-n^(2)+3n-2}d)/(2a+{4n^(2)-2n-4n^(2)+6n-2}d)`
`=(2a(2n+1)+(8n^(2)-2)d)/(2a+(4n-2)d)`
`=(2(2n+1){a+(2n-1)d})/(2{a+(2n-1)d})=2n+1`