The sum of n terms of two arithmetic progressions are in the ratio `(3n + 8) : (7n + 15)`. Find the ratio of their `12^(t h)`terms.

Options

A. `16:7`
B. `7:16`
C. `74:169`
D. none of these

Answer

B

Explanation

If the ratio of the sums of n terms two arithmetic progressions is given, then the ratio of their `n^(th)` terms is obtained by replacing n by (2n-1) in the ratio of the sums of n terms.
Here, we have
Ratio of the terms `=(3n+8)/(7n+15)`
`:." Ratio of "n^(th)" terms"=(3(2n-1)+8)/(7(2n-1)+15)=(6n+15)/(14n+8)`
`rArr" Ratio of "12^(th)" terms "=(6xx12+5)/(14xx12+8)=(77)/(176)=(7)/(16)`