Question

find the sum of first 31 terms of an AP whose nth term is given by 3+2/3n

Answer 1

The sum is 80/3.

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Answer 2

Answer:

The sum of first 31 terms of an AP is $\frac{1271}{3}$.

Step-by-step explanation:

It is given that  nth term is

$a_n=3+\frac{2}{3}n$

First term of the AP is

$a__1=3+\frac{2}{3}(1)=\frac{11}{3}$

Second term of the AP is

$a__2=3+\frac{2}{3}(2)=\frac{13}{3}$

$d=a_2-a_1=\frac{13}{3}-\frac{11}{3}=\frac{2}{3}$

The common difference is 2/3.

The sum of first 31 terms of an AP is

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

$S_{31}=\frac{31}{2}[2(\frac{11}{3})+(31-1)(\frac{2}{3})]$

$S_{31}=\frac{31}{2}[(\frac{22}{3})+30(\frac{2}{3})]$

$S_{31}=\frac{31}{2}(\frac{22+60}{3})$

$S_{31}=\frac{31}{2}(\frac{82}{3})$

$S_{31}=\frac{1271}{3}$

Therefore the sum of first 31 terms of an AP is $\frac{1271}{3}$.