Find the sum of the first 15 terms of each of the following sequences having nth term as$x_n = 6 - n$

Given:

nth term of an A.P. is given by $x_n = 6 - n$.

To do:

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

Solution:

Here,

\( x_{n}=6-n \)

Number of terms \( =15 \)

\( x_{1}=6-1=5 \)
\( x_{2}=6-2=4 \)
\( \therefore \) First term \( (a)=5 \) and \( (d)=4-5=-1 \)

We know that,

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

$=\frac{15}{2}[2 \times 5+(15-1) \times(-1)]$
$=\frac{15}{2}[10+14(-1)]=\frac{15}{2}[10-14]$
$=\frac{15}{2} \times(-4)=15(-2)=-30$
The sum of the first 15 terms is $-30$.

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Updated on: 10-Oct-2022

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