The sum of n terms of two arithmetic progressions are in the ratio (3n+8) : (7n+15) Find the ratio - Maths -

Question

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

Lalit Mehra · Accepted Answer

Let a1, a2 and d1, d2 be the first terms and common difference of two A P 's Given, $\frac{a_{1} + (n - 1) d_{1}}{a_{2} + (n - 1) d_{2}} = \frac{3 n + 8}{7 n + 15}$ Putting n = 12, we get $\frac{a_{1} + 11 d_{1}}{a_{2} + 11 d_{2}} = \frac{3 \times 12 + 8}{7 \times 12 + 15} ∴ \frac{12^{th} term of one A P}{12^{th} term of other A P} = \frac{36 + 8}{84 + 15} = \frac{44}{99} = \frac{4}{9}$ Thus, the ratio of their 12th terms is 4 : 9

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

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