$3 + 5 + 7 + . . . . . . . . . . u p t o n t e r m s = 224$ find $n$

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Solution

$3 + 5 + 7 + . . .$

The series is in AP with $a = 3, d = 2$

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

$S_{n} = \frac{n}{2} (2 (3) + (n - 1) 2) = 224$
$6 n + 2 n^{2} - 2 n = 448$

$n^{2} + 2 n - 224 = 0$
$n^{2} + 16 n - 14 n - 224 = 0$
$n (n + 16) - 14 (n + 16) = 0$
$(n + 16) (n - 14) = 0$
$n = 14, - 16$

$⟹ n = 14$

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