[24] The sum of how many terms of the series 6 + 12 + 18 + 24 + ... is 1800 ?

For Competitive Exams

The sum of how many terms of the series 6 + 12 + 18 + 24 + ... is 1800 ?

Explanation:

This is an A.P. in which $ a $ = 6, $ d $ = 6 and S_n = 1800

Then,$ \dfrac{n}{2} $[2$ a $ +$\left ( n - 1\right)$$ d $] = 1800

$\Rightarrow$ $ \dfrac{n}{2} $[2 x 6 + $\left( n - 1\right)$ x 6] = 1800

$\Rightarrow$ 3$ n $ $\left(n + 1\right)$ = 1800

$\Rightarrow n \left(n + 1\right)$ = 600

$\Rightarrow n $² + $ n $ - 600 = 0

$\Rightarrow n $² + 25$ n $ - 24$ n $ - 600 = 0

$\Rightarrow n $$\left( n + 25\right)$ - 24$\left( n + 25\right)$ = 0

$\Rightarrow$ $\left( n + 25\right)$$\left( n - 24\right)$ = 0

$\Rightarrow n $ = 24

Number of terms = 24.

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