If the $n^{t h}$ term of an A.P be $(2 n - 1)$ , then the sum of its first $n$ terms will be

n^{n} - 1

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(2 n - 1)^{2}

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n^{2}

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n^{2} + 1

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Solution

The correct option is C $n^{2}$
Given, $a_{n} = 2 n - 1$
if $n = 1$ , then
$a_{1} = 2 (1) - 1 = 1$
$S_{n} = \frac{n}{2} [f i r s t t e r m + l a s t t e r m]$
= $\frac{n}{2} (a_{1} + a_{n})$
$\Rightarrow S_{n} = \frac{n}{2} (1 + 2 n - 1)$
$\Rightarrow S_{n} = n^{2}$

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n^{t h}

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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).

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