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# Find the sum of the first 25 terms of an A.P. whose nth term is given by $a_n = 2 â€“ 3n$.

Given:

nth term of an A.P. is given by $a_n = 2 – 3n$.

To do:

We have to find the sum of the first 25 terms.

Solution:

Here,

\( a_{n}=2-3n \)

Number of terms \( =25 \)

\( a_{1}=a=2-3 \times 1=2-3=-1 \)

\( a_{2}=2-3 \times 2=2-6=-4 \)

\( \therefore d=a_{2}-a_{1}=-4-(-1)=-4+1=-3 \)

We know that,

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

\( S_{25}=\frac{25}{2}[2 a+(25-1) d] \)

\( =\frac{25}{2}[2 \times (-1)+(25-1) \times (-3)] \)

\( =\frac{25}{2}[-2+24 \times (-3)]=\frac{25}{2}[-2-72] \)

\( =\frac{25}{2} \times (-74)=25 \times (-37)=-925 \)

The sum of the first 25 terms is $-925$.â€Š

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