Find the sum of the first 15 terms of each of the following sequences having nth term as$y_n = 9 – 5n$

Given:

nth term of an A.P. is given by $y_n = 9 – 5n$.

To do:

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

Solution:

Here,

\( y_{n}=9-5 n \)

Number of terms \( =15 \)

\( y_{1}=9-5 \times 1=9-5=4 \)
\( y_{2}=9-5 \times 2=9-10=-1 \)
\( \therefore \) First term \( (a)=4 \) and common difference \( (d)=y_{2}-{y}_{1}=-1-4=-5 \)

We know that,

\( \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d] \)
\( \mathrm{S}_{15}=\frac{15}{2}[2 a+(15-1) d] \)
\( =\frac{15}{2}[2 \times 4+(15-1)(-5)] \) 

 \( =\frac{15}{2}[8+14(-5)] \)

 \( =\frac{15}{2}(8-70) \) 

\( =\frac{15}{2}(-62) \) 

\( =15(-31) \) 

\( =-465 \)

The sum of the first 15 terms is $-465$.