Question

The sum of n terms of two arithmetic series are in the ratio $2n + 3 : 6n + 5$, then the ratio of their 13th terms is

• A. $53 : 155$
• B. $27 : 87$
• C. $29 : 83$
• D. $31 : 89$

Answer 1

The Correct Option is A

Sum of an A.P. is given by $S_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$ where 'a' is the first term and 'd' is the common difference of $A.P.$ Let $S_{n_1}$ be the sum of n terms of $I^{st}\, A.P.$ and $S_{n_2}$ be the sum of n terms of $II^{nd}\,A.P.$ Given that the sum of n terms of two arithmetic series is in the ratio $2n + 3 : 6n + 5$ $\Rightarrow \frac{S_{n_1}}{S_{n_2}}=\frac{2n+3}{6n+5}\,...\left(i\right)$ $\Rightarrow S_{n_1}=\frac{n}{2}\left[2a_{1}+\left(n-1\right)d_{1}\right]=2n+3$ and $ S_{n_1}=\frac{n}{2}\left[2a_{2}+\left(n-1\right)d_{2}\right]=6n+5 $From E (i) , we get $\frac{S_{n_1}}{S_{n_1}}=\frac{\frac{n}{2}\left[2a_{1}+\left(n-1\right)d_{1}\right]}{\frac{n}{2}\left[2a_{2}+\left(n-1\right)d_{2}\right]}=\frac{2n+3}{6n+5}$ $\Rightarrow \frac{2a_{1}+\left(n-1\right)d_{1}}{2a_{2}+\left(n-1\right)d_{2}}=\frac{2n+3}{6n+5}$ For $a = 13, n = 2a - 1 = 2 ? 13 - 1 = 25$ $\therefore \frac{2a_{1}+\left(25-1\right)d_{1}}{2a_{2}+\left(25-1\right)d_{2}}=\frac{53}{155} \Rightarrow \frac{a_{1}+12d_{1}}{a_{2}+12d_{2}}=\frac{53}{155}$