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

n^{2} - 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 B $n^{2}$
$a_{n} = (2 n - 1)$
$\Rightarrow$ $a_{1} = 2 \times 1 - 1 = 1$
$\Rightarrow$ $a_{2} = 2 \times 2 - 1$
$= 4 - 1$
$= 3$
$\Rightarrow$ $d = a_{1} - a_{1} = 3 - 1$
$∴$ $d = 2$
$\Rightarrow$ $S_{1} = \frac{n}{2} [2 a_{1} + (n - 1) d]$

$= \frac{n}{2} [2 (1) + (n - 1) 2]$

$= \frac{n}{2} [2 + 2 n - 2]$

$= \frac{n}{2} \times 2 n$

$= n^{2}$

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