The $n^{t h}$ term of an $A P$ is $a = 2 = n + 1$ , find its sum.

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Solution

Given,

$a = 2$

$n + 1 = 2 \Rightarrow n = 1$

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$S_{1} = \frac{1}{2} [2 (2) + (1 - 1) d]$

$S_{1} = \frac{4}{2}$

$S_{1} = 2$

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The nth term of an AP is a=2=n+1, find its sum.

Given,a=2n+1=2⇒n=1Sn=n2[2a+(n−1)d]S1=12[2(2)+(1−1)d]S1=42S1=2

The $n^{t h}$ term of an $A P$ is $a = 2 = n + 1$ , find its sum.

Given,

$a = 2$

$n + 1 = 2 \Rightarrow n = 1$

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$S_{1} = \frac{1}{2} [2 (2) + (1 - 1) d]$

$S_{1} = \frac{4}{2}$

$S_{1} = 2$