If Sn denotes the sum of first n terms of an A-P- such that Sm-Sn-m2-n2- then am-an-

The correct option is D 2m 1/2n 1 S_n/S_n = m^2/n^2 ⇒m/2{2a+(m 1)d}/m/2{2a+(n 1)d} = m^2/n^2 ⇒{2a+(m 1)d}/{2a+(n 1)d} = m/n ⇒ 2an + ndm nd=2am + nmd md ⇒ 2an ...

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Standard XII

Mathematics

nth Term of A.P

If Sn denotes...

Question

If $S_{n}$ denotes the sum of first n terms of an A.P. $< a_{n} >$ such that $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$ , then $\frac{a_{m}}{a_{n}} =$

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

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

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

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$\frac{m + 1}{n + 1}$

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Solution

The correct option is B
$\frac{2 m - 1}{2 n - 1}$

$\frac{S_{n}}{S_{n}} = \frac{m^{2}}{n^{2}}$

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

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

$\Rightarrow 2 a n + n d m - n d = 2 a m + n m d - m d$

$\Rightarrow 2 a n - 2 a m - n d + m d = 0$

$\Rightarrow 2 a (n - m) - d (n - m) = 0$

$\Rightarrow 2 a (n - m) - d (n - m)$

$\Rightarrow d = 2 a \dots \dots (1)$

Ratio of $\frac{a_{m}}{a_{n}} = \frac{a + (m - 1) d}{a + (n - 1) d}$

$\Rightarrow \frac{a_{m}}{a_{n}} = \frac{a + (m - 1) 2 a}{a + (n - 1) 2 a}$ From (1)

$= \frac{a + 2 a m - 2 a}{a + 2 a n - 2 a}$

$= \frac{2 a m - a}{2 a m - a} = \frac{(2 m - 1)}{(2 n - 1)}$

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If Sn denotes the sum of first n terms of an A.P. <an> such that SmSn=m2n2, then aman=

If $S_{n}$ denotes the sum of first n terms of an A.P. $< a_{n} >$ such that $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$ , then $\frac{a_{m}}{a_{n}} =$