Question

For an A.P if $\frac{S_m}{S_n}=\frac{m^2}{n^2}$, then $\frac{t_m}{t_n}=$______

• A. $\frac{2n-1}{2m+1}$
• B. $\frac{2m-1}{2n-1}$
• C. $\frac{m}{n}$
• D. $\frac{n}{m}$

Answer 1

Answer: (B)

$\frac{2m-1}{2n-1}$

Let $a$ be the first term and $d$ be the common difference of the given $A.P$

Now,

$\frac{S_m}{S_n}=\frac{m^2}{n^2}$

$\frac{\frac{m}{2}\left|2a+\left(m-1\right)d\right|}{\frac{n}{2}\left|2a+\left(n-1\right)d\right|}=\frac{m^2}{n^2}$ [ cancle $2$ both numerator and denominator and also cancle $m$ and $n$ both sides]

$\frac{2a+\left(m-1\right)d}{2a+\left(n-1\right)d}=\frac{m}{n}-----\left(i\right)$

Now,

$\frac{t_m}{t_n}=\frac{a+\left(m-1\right)d}{a+\left(n-1\right)d}$

$=\frac{2a+2\left(m-1\right)d}{2a+2\left(n-1\right)d}$

$=\frac{2a+\left(2m-2\right)d}{2a+\left(2n-2\right)d}$

$\therefore \frac{t_m}{t_n}==\frac{2a+\left[2\left(m-1\right)d\right]}{2a+\left[2\left(n-1\right)d\right]}-----\left(ii\right)$

Put $2m-1$ in place of $p$ and $(2n-1)$ in place of $q$ in $\left(i\right)$ we get,

$\frac{2a+\left[2\left(m-1\right)d\right]}{2a+\left[2\left(n-1\right)d\right]}=\frac{2m-1}{2n-1}-----\left(iii\right)$

From $\left(ii\right)$ and $\left(iii\right)$ we have

$\frac{t_m}{t_n}=\frac{2m-1}{2n-1}$

Hence, the option $\left(B\right)$ is correct.