Question

If the ratio of sum of n terms of two different AP’s is (7n – 4) : (5n + 3), then the ratio of their mth term is equal to

• A. $\frac{(7m-4)}{(4m+3)}$
• B. $\frac{(14m-11)}{(10m-2)}$
• C. $\frac{(17m+9)}{(12m-5)}$
• D. $\frac{(5m+3)}{(7m-4)}$

Answer 1

The correct option is B $\frac{(14m-11)}{(10m-2)}$

Let the first terms of the two AP’s with n_{th} terms $a_n$ and $a_{n\text{'}}$ be a and a′, and common difference be d and d′, respectively.
Now, Sum of n terms of first AP, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Sum of n terms of Second AP, $S_n^{\prime }=\frac{n}{2}[2a^{\prime }+(n-1\left)d^{\prime }\right]$
Given: $\frac{S_n}{S_n^{\prime }}=\frac{\frac{n}{2}[2a+(n-1\left)d\right]}{\frac{n}{2}[2a^{\prime }+(n-1\left)d^{\prime }\right]}=\frac{7n-4}{5n+3}$
$\Rightarrow \frac{a+\frac{(n-1)}{2}d}{a^{\prime }+\frac{(n-1)}{2}{d}^{\prime }}=\frac{7n-4}{5n+3}$ ........(i)
Substituting $\frac{n-1}{2}=m-1$, i.e., n = 2m – 1 in (i), we get
$\frac{a+(m-1)d}{a^{\prime }+(m-1)d^{\prime }}=\frac{7(2m-1)-4}{5(2m-1)+3}=\frac{14m-11}{10m-2}$
$\Rightarrow \frac{a_m}{a_m^{\prime }}=\frac{14m-11}{10m-2}$
Hence, the correct answer is option (2).