The nth term of an A.P. is given by $(-4n + 15)$. Find the sum of first 20 terms of this A.P.

Given:

The nth term of an A.P. is given by $(-4n + 15)$.

To do:

We have to find the sum of first 20 terms of the A.P.

Solution:

\( n^{\text {th }} \) term of the given A.P. is \( a_{n}=-4n+15 \).

First term \( =a_{1}=-4 \times 1+15=-4+15=11 \)

\( a_{2}=-4 \times 2+15=-8+15=7 \)

\( \therefore d=a_{2}-a_{1}=7-(11)=-4 \)

We know that,

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

\( S_{20}=\frac{20}{2}[2 \times 11+(20-1)(-4)] \)

\( =10[22+19(-4)]=10[22-76] \)

\( =10 \times (-54)=-540 \)

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Updated on: 10-Oct-2022

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