If the sums of n terms of two arithmetic progressions are in the ration 3n + 55n + 7, then their nth terms are in the ration

Answer: Option B

In first A.P. let its first term be a₁ and common difference d₁
and in second A.P., first term be a₂ and common difference d₂, then
$$\begin{array}{rl}& \frac{S_n}{S_n}=\frac{\frac{n}{2}\left[2a_1+\left(n-1\right)d_1\right]}{\frac{n}{2}\left[2a_2+\left(n-1\right)d_2\right]}\\ & =\frac{2a_1+\left(n-1\right)d_1}{2a_2+\left(n-1\right)d_2}\end{array}$$
$$\therefore \frac{2a_1+\left(n-1\right)d_1}{2a_2+\left(n-1\right)d_2}=\frac{3n+5}{5n+7}$$
$$\text{Substituting n = 2n - 1, then}$$
$$\frac{2a_1+\left(2n-2\right)d_1}{2a_2+\left(2n-2\right)d_2}=$$     $$\frac{3\left(2n-1\right)+5}{5\left(2n-1\right)+7}$$
$$\Rightarrow \frac{a_1+\left(n-1\right)d_1}{a_2+\left(n-1\right)d_2}=$$     $$\frac{6n-3+5}{10n-5+7}$$    (Dividing by 2)
$$\begin{array}{rl}& \Rightarrow \frac{a_{1n}}{a_{2n}}=\frac{6n+2}{10n+2}\\ & \Rightarrow \frac{a_{1n}}{a_{2n}}=\frac{3n+1}{5n+1}\end{array}$$