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

Explanation

Let `a` is the first term and `d_1` is the common difference of the first AP.

Let `b` is the first term and `d_2` is the common difference of the second AP.

Then,

`(n/2(2a+(n-1)d_1))/(n/2(2b+(n-1)d_2)) = (3n+8)/(7n+15)`

`=>(2a+(n-1)d_1)/(2b+(n-1)d_2) =(k(3n+8))/(k(7n+15))`

Comparing numerators for both sides,

`=>2a-d_1 = 8k, d_1 = 3k`

`=> a = 11/2k`

...