Find the sum of $20$ terms of A.P whose $n^{t h}$ term is $3 n - 1$ .

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Solution

Given: $t_{n} = 3 n - 1$
$∴ t_{1} = 3 \times 1 - 1 = 3 - 1 = 2$
$t_{2} = 3 \times 2 - 1 = 6 - 1 = 5$
$t_{1} = 3 \times 3 - 1 = 9 - 1 = 8$ and so on
$∴, a = 2, d = 5 - 2 = 3, n = 20$
$S_{20} = \frac{20}{2} [2 \times 2 + (20 - 1) 3]$
$= 10 [4 + 57]$
$= 10 \times 61 = 610$

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