It is known that the equality $\frac{s_{m}}{s_{n}} = \frac{m^{2}}{n^{2}}$ holds for a certain arithmetic progession and for a certain pair of natural numbers m& n If $a_{m}$ & $a_{n}$ denotes the $m^{th}$ and $n^{th}$ terms of the A.P. respectively, then $a_{m} / a_{n}$ equals

\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 A $\frac{2 m - 1}{2 n - 1}$
let a = first term
d= common difference
$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$
$\frac{\frac{m}{2} (2 a + (m + 1) d)}{\frac{n}{2} (2 a + (n - 1) d)} = \frac{m^{2}}{n^{2}}$
$\frac{2 a + (m - 1) d}{2 a + (n - 1) d} = \frac{m}{n}$
$20 n + (m n - n) d = 2 a m + (m n - m) d$
$2 a (n - m) + (m n - n - m n + m) d = 0$
$2 a (n - m) + (m - n) d = 0$
$(m - n) (d - 2 a) = 0$
$d = 2 a$
$\begin{matrix} \frac{a_{m}}{a_{n}} & = \frac{a + (m - 1) d}{a + (n - 1) d} = \frac{a + (m - 1) 2 a}{a + (n - 1) 2 a} = \frac{2 m - 1}{2 n - 1} option A & is correct. \end{matrix}$

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Similar questions

If the ratio of sum of m terms and n terms of an A.P. is $m^{2}$ : $n^{2}$ , then the ratio of its $m^{t h}$ and $n^{t h}$ terms will be :

If the ratio of sum of m terms and n terms of an A.P. is $m^{2}$ : $n^{2}$ , then the ratio of its $m^{t h}$ and $n^{t h}$ terms will be

If the ratio of the sum m terms and n terms of an A.P. are m²:n² . Prove that the ratio of its m^th and n^th terms will be (2m-1):(2n-1).

m^{2} : n^{2}

m^{t h}

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A . P .

m^{2} : n^{2}

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2 m - 1 : 2 n - 1

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