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Question: If ratio of sum of n terms of two AP's be 7n + 1 : 4n + 27 then ratio of their 11<sup>th</sup> term...

If ratio of sum of n terms of two AP's be

7n + 1 : 4n + 27 then ratio of their 11th term =

A

2 : 3

B

3 : 4

C

4 : 3

D

5 : 6

Answer

4 : 3

Explanation

Solution

SnSn′\frac { S _ { n } } { S _ { n } ^ { \prime } } = 7n+14n+27\frac { 7 n + 1 } { 4 n + 27 } ̃ tntn′\frac { \mathrm { t } _ { \mathrm { n } } } { \mathrm { t } _ { \mathrm { n } } ^ { \prime } } = 7(2n−1)+14(2n−1)+27\frac { 7 ( 2 n - 1 ) + 1 } { 4 ( 2 n - 1 ) + 27 } =

=154−688+23\frac { 154 - 6 } { 88 + 23 }=148111\frac { 148 } { 111 }= 43\frac { 4 } { 3 }