 ### 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 B

### 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}

A. 22
B. 25
C. 23
D. 24

A. 5
B. 6
C. 4
D. 3

A. -45
B. -55
C. -50
D. 0