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

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

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Sum of n terms is 1800
first term (a) = 6
common difference (d) = 6
by formula, sum of n terms = n/2[2a+(n-1)d]

\begin{array}{l} 1800=n / 2[12+(n-1) 6] \\ 3600=n[12+6 n-6] \\ 3600=6 n+6 n^{2} \\ 6 n^{2}+6 n-3600=0 \\ n^{2}+n-600=0 \\ n^{2}+25 n-24 n-600=0 \\ n(n+25)-24(n+25)=0 \\ (n+25)(n-24)=0 \\ so, n=24 \end{array}

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lovekush

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