If the sums of n terms of two arithmetic progressions are in the ration $\Large\frac{3 n+5}{5 n+7}$, then their nth terms are in the ration.
A. $\begin{aligned}\frac{3 n-1}{5 n-1}\end{aligned}$ B. $\begin{aligned}\frac{3 n+1}{5 n+1}\end{aligned}$ C. $\begin{aligned}\frac{5 n+1}{3 n+1}\end{aligned}$ D. $\begin{aligned}\frac{5 n-1}{3 n-1}\end{aligned}$ Answer: Option BShow Answer
Solution(By Apex Team)
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,
$\begin{aligned}\frac{S_n}{S_n}&=\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}\\
&=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}\\
&\therefore\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{3n+5}{5n+7}\end{aligned}$
Substituting n = 2n – 1, then
$\begin{aligned}&\frac{2a_1+(2n-2)d_1}{2a_2+(2n-2)d_2}=\frac{3(2n-1)+5}{5(2n-1)+7}\\
&\Rightarrow\frac{a_1+(n-1)d_1}{a_2+(n-1)d_2}=\frac{6n-3+5}{10n-5+7}\text{ (Dividing by 2) }\\
&\Rightarrow\frac{a_{1n}}{a_{2n}}=\frac{6n+2}{10n+2}\\
&\Rightarrow\frac{a_{1n}}{a_{2n}}=\frac{3n+1}{5n+1}\end{aligned}$
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