Find the sum of the first 11 terms of the A.P.: $2, 6, 10, 14,…..$

Given:

Given A.P. is $2, 6, 10, 14,…..$

To do:

We have to find the sum of the first 11 terms of the A.P.
Solution:

Here,

First term \( (a)=2 \) and common difference \( (d)=6-2=4 \)

We know that,
\( \therefore \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d] \)
$\mathrm{S}_{11}=\frac{11}{2}[2 \times 2+(11-1) \times 4]$
$=\frac{11}{2}[4+10 \times 4]$

$=\frac{11}{2}(4+40)$
$=\frac{11}{2} \times 44$

$=11 \times 22$

$=242$

The sum of the first 11 terms of the given A.P. is $242$.

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Updated on: 10-Oct-2022

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