Question

5+10+15+...+100 write the series using summation notation

Answer 1

The Solution.

To determine that the series is an arithmetic progression,

`\begin{gathered} T_{2_{}}-T_1=T_3-T_2=d \\ \text{Where d = common difference} \end{gathered}`
`d=10-5=15-10=5`

The sum of n terms of an arithmetic progression is given as

`\begin{gathered} S_n=(n)/(2)(a+l) \\ \text{Where S}_n=\sum ^(\square)_(\square) \\ n=n\text{ umber of terms}=\text{?} \\ a=\text{first term=5} \\ l=\text{last term=100} \end{gathered}`

But we need to first find the number of terms (n), by using the formula below:

`\begin{gathered} l=a+(n-1)d \\ \text{Where a = 5, l=100, d = 5 and n =?} \end{gathered}`

Substituting the values, we get

`\begin{gathered} 100=5+(n-1)5 \\ 100=5+5n-5 \\ 100=5n \\ \text{Dviding both sides by 5, we get} \\ n=(100)/(5)=20 \end{gathered}`

Substituting into the formula for finding the sum of terms of the series, we get

`\begin{gathered} S_(20)=(20)/(2)(5+100) \\ \text{ } \\ \text{ = 10(105) = 1050} \end{gathered}`

Therefore, the correct answer is 1050.