Question

Sum of the first n terms of an arithmetic sequence

Answer 1

Given

$\begin{gathered} a)\text{ }-1+3+7+...+551 \\ b)\text{ }\sum_{i\mathop{=}1}^{79}(2i+6) \end{gathered}$

To find the sum of the arithmetic sequence.

Explanation:

It is given that,

$\begin{gathered} a)\text{ }-1+3+7+...+551 \\ b)\text{ }\sum_{i\mathop{=}1}^{79}(2i+6) \end{gathered}$

That implies,

a) Here, the first term, the last t and common difference is,

$\begin{gathered} a=-1,d=3-(-1)=3+1=4,l=551 \\ \Rightarrow a=-1,d=4,l=551 \end{gathered}$

And,

$\begin{gathered} n=\frac{l-a}{d}+1 \\ =\frac{551-(-1)}{4}+1 \\ =\frac{552}{4}+1 \\ =138+1 \\ =139 \end{gathered}$

Then, the sum is given by,

$\begin{gathered} S_n=\frac{n}{2}(a+l) \\ =\frac{139}{2}(-1+551) \\ =\frac{139}{2}\times550 \\ =139\times275 \\ =38225 \end{gathered}$

Hence, the sum is 38225.

Also,

b) It is given that,

$\sum_{i=1}^{79}(2i+6)=8,10,12,...,164$

Here, the first term, common difference and the last term are,

$\begin{gathered} a=8,d=10-8=2,l=164 \\ a=8,d=2,l=164 \end{gathered}$

And,

$n=79$

Then, the sum is given by,

$\begin{gathered} S_n=\frac{n}{2}(a+l) \\ S_{79}=\frac{79}{2}(8+164) \\ =\frac{79}{2}\times172 \\ =79\times86 \\ =6794 \end{gathered}$

Hence, the sum is 6794.