Find the sum of $2, 4, 6, 8,,, . . . . . . . .2 n$

n (n + 1)

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n^{2}

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n (n - 1)

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n^{2} + n

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Solution

The correct option is A $n (n + 1)$
The given sequence is $2, 4, 6, 8, . . . . . 2 n$
$∴ a = 2, d = 2$ & $a_{k} = 2 n$
$∴ 2 n = a + (k - 1) d$
$\Rightarrow 2 n = 2 + (k - 1) 2$
$\Rightarrow k = n$
Sum of $n$ terms $= \frac{k}{2} (2 a + (k - 1) d)$
$= \frac{n}{2} \times [2 (2) + (n - 1) (2)]$
$= n (n + 1)$

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