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

^{th}terms are:
(1) 9/4 (2) 16/7
(3) 7/16 (4) 4/9 (5) none of these

Answer follows here:

**Solution:**
Let the first term and
common difference of the first A.P be a

_{1}and d_{1}. And let the first term and common difference of the second A.P be a_{2}and d_{2}.
(Sum to n terms of first
A.P)/( Sum to n terms of second A.P) = (3n+8)/(7n+15)

n/2 * [2a

_{1}+(n-1)d_{1}] / n/2 * [2a_{2}+(n-1)d_{2}] = (3n+8)/(7n+15)
[2a

_{1}+(n-1)d_{1}]_{ }/ [2a_{2}+(n-1)d_{2}] = (3n+8)/(7n+15) ----(1)
We need (12

^{th}term of first A.P)/( 12^{th}term of second A.P) = [a_{1}+11d_{1}]_{ }/ [a_{2}+11d_{2}]
Substituting, n = 23 in
(1),

[2a

_{1}+22d_{1}]_{ }/ [2a_{2}+22d_{2}] = (69+8)/(161+15) = 77/176 = 7/16
=> [a

_{1}+11d_{1}]_{ }/ [a_{2}+11d_{2}] = 7/16**Answer (3)**
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