Progressions-6 (NCERT)

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 a₁ and d₁. And let the first term and common difference of the second A.P be a₂ and d₂.

(Sum to n terms of first A.P)/( Sum to n terms of second A.P) = (3n+8)/(7n+15)

n/2 * [2a₁+(n-1)d₁] / n/2 * [2a₂+(n-1)d₂] = (3n+8)/(7n+15)

[2a₁+(n-1)d₁] / [2a₂+(n-1)d₂] = (3n+8)/(7n+15)                ----(1)

We need (12^th term of first A.P)/( 12^th term of second A.P) = [a₁+11d₁] / [a₂+11d₂]

Substituting, n = 23 in (1),

[2a₁+22d₁] / [2a₂+22d₂] = (69+8)/(161+15) = 77/176 = 7/16

=> [a₁+11d₁] / [a₂+11d₂] = 7/16

Answer (3)