The ratio of the sums to first n terms of two APs is (3n+1):(5n+8) Find the ratio of their - Maths - Circles

Question

The ratio of the sums to first n terms of two APs is
(3n+1):(5n+8) Find the ratio of their 13th terms

Lalit Mehra · Accepted Answer

Let a1, a2 and d1, d2 be the first terms and common difference of two A P 's Suppose $S_{n_{1}}$ and $S_{n_{2}}$ denote the sum of n terms of two A P 's $\frac{S_{n_{1}}}{S_{n_{2}}} = \frac{3 n + 1}{5 n + 8} (Given) ∴ \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{3 n + 1}{5 n + 8} \Rightarrow \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{3 n + 1}{5 n + 8} \Rightarrow \frac{a_{1} + (\frac{n - 1}{2}) d_{1}}{a_{2} + (\frac{n - 1}{2}) d_{2}} = \frac{3 n + 1}{5 n + 8} Putting \frac{n - 1}{2} = 12, we get \frac{a_{1} + 12 d_{1}}{a_{2} + 12 d_{2}} = \frac{3 \times 25 + 1}{5 \times 25 + 8} (∵ \frac{n - 1}{2} = 12 \Rightarrow n - 1 = 24 \Rightarrow n = 25) \Rightarrow \frac{a_{13_{1}}}{a_{13_{2}}} = \frac{75 + 1}{125 + 8} = \frac{76}{133} \Rightarrow a_{13_{1}} : a_{13_{2}} = 76 : 133$ Thus, the ratio of their 13th terms is 76 : 133

The ratio of the sums to first n terms of two APs is (3n+1):(5n+8). Find the ratio of their 13th terms.