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

Answer: (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 $I{I}^{nd}A.P.$
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
$S_{n_1}=\frac{n}{2}\left[2a_2+\left(n-1\right)d_2\right]=6n+5<br/>$From Eq. (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 $a=13,n=2a-1=2\times 13-1=25$
$\therefore \frac{2a_1+\left(25-1\right)d_1}{2a_2+\left(25-1\right)d_2}=\frac{53}{155}\Rightarrow \frac{a_1+12d_1}{a_2+12d_2}=\frac{53}{155}$