Find the sum of the following arithmetic progressions:$ a+b, a-b, a-3 b, \ldots $ to 22 terms

Given:

Given A.P. is \( a+b, a-b, a-3 b, \ldots \)

To do:

We have to find the sum of the given A.P. to 22 terms.

Solution:

Here,

\( a=a+b, d=a-b-a-b=-2 b \) and \( n=22 \)

We know that,

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

\( \therefore S_{22}=\frac{22}{2}[2 \times(a+b)+(22-1)(-2 b)] \)

\( =11[2 a+2 b+21 \times(-2 b)] \)

\( =11[2 a+2 b-42 b] \)

\( =11[2 a-40 b] \)

\( =22 a-440 b \)

The sum of the given A.P. to 22 terms is $22a-440b$.

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Updated on: 10-Oct-2022

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