Find $t n$ and $S n$ for an $A . P . 4, 9, 14, 19, . . . . . . . . . .$

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Solution

First term $= t_{1} = a = 4$
Second term $= t_{2} = 9$
Common difference $= d = t_{2} - t_{1} = 9 - 4 = 5$
$t_{n} = a + (n - 1) d$
$∴ t_{n} = 4 + (n - 1) 5 = 4 + 5 n - 5 = 5 n - 1$
$∴ l = t_{n} = 5 n - 1$
$S_{n} = \frac{n}{2} (a + l)$
$= \frac{n}{2} (4 + 5 n - 1)$
$= \frac{n}{2} (5 n + 3)$
$∴ S_{n} = \frac{n}{2} (5 n + 3)$

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