Find the sum of the first $22$ terms of an AP, whose first term is $4$ and the common difference is $\frac{4}{3}$ .

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Solution

Given that, $a = 4$ and $d = \frac{4}{3}$
We have $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{22} = (\frac{22}{2}) [(2) (4) + (22 - 1) (\frac{4}{3})]$
$= (11) (8 + 28) = 396$

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Find the sum of the first 22 terms of an AP, whose first term is 4 and the common difference is 43.

Given that, a=4 and d=43We have Sn=n2[2a+(n−1)d]S22=(222)[(2)(4)+(22−1)(43)] =(11)(8+28)=396

Find the sum of the first $22$ terms of an AP, whose first term is $4$ and the common difference is $\frac{4}{3}$ .

Given that, $a = 4$ and $d = \frac{4}{3}$
We have $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{22} = (\frac{22}{2}) [(2) (4) + (22 - 1) (\frac{4}{3})]$
$= (11) (8 + 28) = 396$