Question

In an A. P., $S_m:S_n=m^2:n^2$ . The ratio of $p^{2}th$ term to $q^{2}th$ term is:

• A. $\frac{2p^2+1}{2q^2+1}$
• B. $\frac{2p^2-1}{2q^2-1}$
• C. $\frac{2p-1}{2q-1}$
• D. $\frac{p^2-2}{q^2-2}$

Answer 1

The correct option is B

$\frac{2p^2-1}{2q^2-1}$

Explanation for the correct option:

Step 1: Simplify $S_m:S_n=m^2:n^2$

$S_m=\frac{m}{2}\left[2a+\left(m-1\right)d\right]$ [Sum of numbers in A.P]

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

Now,

$\begin{array}{rcl}\frac{{\displaystyle \frac{m}{2}}\left[2a+\left(m-1\right)d\right]}{{\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]}& =& \frac{m^2}{n^2}\\ \Rightarrow \frac{m\left[2a+\left(m-1\right)d\right]}{n\left[2a+\left(n-1\right)d\right]}& =& \frac{m^2}{n^2}\\ \Rightarrow \frac{\left[2a+\left(m-1\right)d\right]}{\left[2a+\left(n-1\right)d\right]}& =& \frac{m}{n}\end{array}$

By taking $2$ common, from numerator and denominator, we get

$\frac{a+\left({\displaystyle \frac{m-1}{2}}\right)d}{a+\left({\displaystyle \frac{n-1}{2}}\right)d}=\begin{array}{l}\frac{m}{n}\end{array}...\left(1\right)$

Step 2: Find the ratio of $p^{2}th$ term to $q^{2}th$ term.

Let $\begin{array}{rcl}\frac{m-1}{2}& =& p^2-1\mathrm{and}\frac{n-1}{2}=q^2-1\end{array}$ then $m=2p^2-1\mathrm{and}n=2q^2-1$

By substituting the value of $m\mathrm{and}n$ in $\left(1\right)$, we get

$\frac{\left[a+\left(p^2-1\right)d\right]}{\left[a+\left(q^2-1\right)d\right]}=\frac{2p^2-1}{2q^2-1}$

The $n^{th}$ term of A.P is $=a+\left(n-1\right)d$. So, $a+\left(p^2-1\right)d$ will be the $p^{2}th$ term and $a+\left(q^2-1\right)d$ will be the $q^{2}th$ term.

$\begin{array}{rcl}\frac{p^{2}th\mathrm{term}}{q^{2}th\mathrm{term}}& =& \frac{2p^2-1}{2q^2-1}\\ \Rightarrow \frac{p^{2}th\mathrm{term}}{q^{2}th\mathrm{term}}& =& \frac{2p^2-1}{2q^2-1}\end{array}$

Hence, option B is correct.