Find the least value of $n$ for which the sum of the series $3 + 6 + 9 + . . . .$ upto $n$ terms exceeds $1000.$

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Solution

Clearly $T_{n} = 3 n$
$∴ S_{n} = 3 n \sum r = 1 r = \frac{3 n [n + 1]}{2} > 1000$
or $n^{2} + n > \frac{2}{3} \cdot 1000$
or ${[n + \frac{1}{2}]}^{2} \geq 666.9 \Rightarrow n + \frac{1}{2} > 25.8$
or $n > 25.8 - 0.5 = 25.3$
Since $n > 25.3$ and $n$ is an integer, therefore least value of $n$ is $26$ .

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