Find the sum of the following arithmetic progressions:$ 41,36,31, \ldots $ to 12 terms

Given:

Given A.P. is \( 41,36,31, \ldots \)

To do:

We have to find the sum of the given A.P. to 12 terms.

Solution:

Here,

\( a=41, d=36-41=-5 \) and \( n=12 \)

We know that,

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

\( \therefore S_{12}=\frac{12}{2}[2 \times 41+(12-1) \times(-5)] \)

\( =6[(82+11 \times(-5)] \)

\( =6[82-55]=6 \times 27=162 \)

The sum of the given A.P. to 12 terms is 162.

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Updated on: 10-Oct-2022

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