Find the sum of the following arithmetic progressions:$ 3, \frac{9}{2}, 6, \frac{15}{2}, \ldots $ to 25 terms

Given:

Given A.P. is \( 3, \frac{9}{2}, 6, \frac{15}{2}, \ldots \)

To do:

We have to find the sum of the given A.P. to 25 terms.
Solution:

Here,

\( a=3, d=\frac{9}{2}-3=\frac{3}{2} \) and \( n=25 \)

We know that,

\( \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d] \)
\( \therefore \mathrm{S}_{25}=\frac{25}{2}\left[2 \times 3+(25-1) \times \frac{3}{2}\right] \)
\( =\frac{25}{2}\left[6+24 \times \frac{3}{2}\right]=\frac{25}{2}[6+36] \)
\( =\frac{25}{2} \times 42=25 \times 21=525 \)

The sum of the given A.P. to 25 terms is 525.

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Updated on: 10-Oct-2022

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