Question

If ${{S}_{n}}$ denotes the sum of first $n$ terms of $A.P.$ $,$ such that $\frac{{{S}_{m}}}{{{S}_{n}}}=\frac{{{m}^{2}}}{{{n}^{2}}},$ then $\frac{{{a}_{m}}}{{{a}_{n}}}$ is equal to

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

Answer 1

Answer: (B)

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

Answer 2

Let a and d be the first term and common difference of an AP respectively, then,
${{S}_{m}}=\frac{m}{2}[2a+(m-1)d]$ and ${{S}_{n}}=\frac{n}{2}[2a+(n-1)d]$
Given, $\frac{{{S}_{m}}}{{{S}_{n}}}=\frac{{{m}^{2}}}{{{n}^{2}}}$
$\Rightarrow$ $\frac{\frac{m}{2}\\{2a+(m-1)d\\}}{\frac{n}{2}\\{2a+(n-1)d\\}}=\frac{{{m}^{2}}}{{{n}^{2}}}$
$\Rightarrow$ $\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$
$\Rightarrow$ $2an+(mn-n)d=2am+(mn-m)d$
$\Rightarrow$ $2a(n-m)+(mn-n-mn+m)d=0$
$\Rightarrow$ $2a(n-m)+(m-n)d=0$
$\Rightarrow$ $(m-n)(d-2a)=0$
$\Rightarrow$ $d=2a$ ..(iii)
$(\because \,\,\,\,m\ne n)$ Now, $\frac{{{a}_{m}}}{{{a}_{n}}}=\frac{a+(m-1)d}{a+(n-1)d}=\frac{a+(m-1)2a}{a+(n-1)2a}$
$=\frac{a(1+2m-2)}{a(2+2n-2)}=\frac{2m-1}{2n-1}$