Question

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

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

Answer 1

The correct option is A

$53:155$

explanation for the correct option:

Find the required ratio.

Given: the sum of $n$ terms of two arithmetic series are in the ratio$2n+3:6n+5$

Let $\frac{S_1}{S_2}=\frac{(2n+3)}{(6n+5)}$

We know that the sum of $n$ terms of the G.P. $S=\frac{n}{2}\left[2a+(n-1)d\right]$

Let $a_1,a_2$ be the first term and $d_1,d_2$ be the common difference

So, $S_1=\frac{n}{2}\left[2a_1+(n-1)d_1\right]$

and $S_2=\frac{n}{2}\left[2a_2+(n-1)d_2\right]$

$$\begin{gathered} \therefore \frac{S_1}{S_2}=\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]} \\ \Rightarrow \frac{(2n+3)}{(6n+5)}=\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]} \\ \Rightarrow \frac{(2n+3)}{(6n+5)}=\frac{\left[2a_1+(n-1)d_1\right]}{\left[2a_2+(n-1)d_2\right]} \\ \Rightarrow \frac{(2n+3)}{(6n+5)}=\frac{\left[a_1+{\displaystyle \frac{1}{2}}(n-1)d_1\right]}{\left[a_2+\frac{1}{2}(n-1)d_2\right]}...\left(i\right) \end{gathered}$$

We also know that, the$n^{th}$ term of an AP $=a+(n-1)d$

Now we need the ratio of their ${13}^{th}$ terms that means, $\frac{(a_1+12d_1)}{(a_2+12d_2)}$

If we put $n=25$ in RHS of $\left(i\right)$ then, we get

​$\frac{(a_1+12d_1)}{(a_2+12d_2)}$

Now put $n=25$ equation $\left(i\right)$, we get

$$\begin{gathered} \frac{\left(2\right(25)+3)}{\left(6\right(25)+5)}=\frac{(a_1+12d_1)}{(a_2+12d_2)} \\ \Rightarrow \frac{(a_1+12d_1)}{(a_2+12d_2)}=\frac{53}{155} \end{gathered}$$

Therefore, the ratio of two arithmetic series of their ${13}^{th}$ terms is $53:155$.

Hence, option (A) is the correct answer.