Find the sum of the first 15 terms of each of the following sequences having nth term as$a_n = 3 + 4n$

Given:

nth term of an A.P. is given by $a_n = 3 + 4n$.

To do:

We have to find the sum of the first 15 terms.

Solution:

Here,

\( a_{n}=3+4 n \)

Number of terms \( =15 \)

\( a_{1}=3+4 \times 1=3+4=7 \) or \( a=7 \)

\( a_{2}=3+4 \times 2=3+8=11 \)
\( \therefore d=a_{2}-a_{1}=11-7=4 \)
We know that,

\( S_{n}=\frac{n}{2}[2 a+(n-1) d] \)
\( S_{15}=\frac{15}{2}[2 a+(15-1) d] \)
\( =\frac{15}{2}[2 \times 7+(15-1) \times 4] \)
\( =\frac{15}{2}[14+14 \times 4]=\frac{15}{2}[14+56] \)
\( =\frac{15}{2} \times 70=15 \times 35=525 \)

The sum of the first 15 terms is $525$. 

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Updated on: 10-Oct-2022

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