Find the sum of the first13 terms of the A.P.: $-6, 0, 6, 12,…..$

Given:

Given A.P. is $-6, 0, 6, 12,…..$

To do:

We have to find the sum of the first 13 terms of the A.P.
Solution:

Here,

First term \( (a)=-6 \) and common difference \( (d)=0-(-6)=6 \)

We know that,
\( \therefore \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d] \)
$\mathrm{S}_{13}=\frac{13}{2}[2 \times (-6)+(13-1) \times 6]$
$=\frac{13}{2}[-12+12 \times 6]$

$=\frac{13}{2}(-12+72)$
$=\frac{13}{2} \times 60$

$=13 \times 30$

$=390$

The sum of the first 13 terms of the given A.P. is $390$.

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Updated on: 10-Oct-2022

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