Find the value of N if
$1 + 4 + 7 + 10 + . . . . . . .$ to n terms $= 590$

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Solution

we have
$1 + 4 + 7 + 10 + . . . . . . .$ to n terms $= 590$
so, here $a = 1$
$d = 3$
$s_{n} = 590$
$\Rightarrow [2 \times 1 + (n - 1) 3] = 590$
$\Rightarrow [2 + 3 n - 3] = 590$
$\Rightarrow n [3 n - 1] = 1180$
$\Rightarrow 3 n^{2} - n - 1180 - 0$
$\Rightarrow 3 n (n - 20) + 59 (n - 20) = 0$
$\Rightarrow (n - 20) (3 n + 59) = 0$
$\Rightarrow n - 20 = 0$
$\Rightarrow n = 20$

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