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

2m-1 : 2n-1

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m:n

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2m+1 : 2n+1

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None

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Solution

The correct option is D 2m-1 : 2n-1
By using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ we have

Given that $\frac{\frac{m}{2} [2 a + (m - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{m^{2}}{n^{2}}$

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

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

$\Rightarrow a (n - m) + \frac{d}{2} [m n - n - m n + m] = 0$

$\Rightarrow a (n - m) + \frac{d}{2} (m - n) = 0$

$\Rightarrow a = \frac{d}{2}$ or $d = 2 a$

Therefore, $\frac{T_{m}}{T_{n}} = \frac{a + (m - 1) d}{a + (n - 1) d}$

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

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

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

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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

m

n

A . P .

m^{2} : n^{2}

m t h

2 m - 1 : 2 n - 1

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

n

m^{2} : n^{2},

m^{t h}

n^{t h}

(2 m - 1) : (2 n - 1)

1

0.

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