Find the sum of the following arithmetic progressions: $ 50,46,42, \ldots $ to 10 terms

Given:

Given A.P. is \( 50,46,42, \ldots \)

To do:

We have to find the sum of the given A.P. to 10 terms.
Solution:

Here, \( a=50, d=46-50=-4 \) and \( n=10 \)
We know that,

\( S_{n}=\frac{n}{2}[2 a+(n-1) d] \)
\( \therefore S_{10}=\frac{10}{2}[2 \times 50+(10-1) \times(-4)] \)
\( =5(100+9 \times(-4)]=5[100-36] \)
\( =5 \times 64=320 \)

The sum of the given A.P. to 10 terms is $320$.

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Updated on: 10-Oct-2022

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