Respuesta :
The Solution.
To determine that the series is an arithmetic progression,
[tex]\begin{gathered} T_{2_{}}-T_1=T_3-T_2=d \\ \text{Where d = common difference} \end{gathered}[/tex][tex]d=10-5=15-10=5[/tex]The sum of n terms of an arithmetic progression is given as
[tex]\begin{gathered} S_n=\frac{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}[/tex]But we need to first find the number of terms (n), by using the formula below:
[tex]\begin{gathered} l=a+(n-1)d \\ \text{Where a = 5, l=100, d = 5 and n =?} \end{gathered}[/tex]Substituting the values, we get
[tex]\begin{gathered} 100=5+(n-1)5 \\ 100=5+5n-5 \\ 100=5n \\ \text{Dviding both sides by 5, we get} \\ n=\frac{100}{5}=20 \end{gathered}[/tex]Substituting into the formula for finding the sum of terms of the series, we get
[tex]\begin{gathered} S_{20}=\frac{20}{2}(5+100) \\ \text{ } \\ \text{ = 10(105) = 1050} \end{gathered}[/tex]Therefore, the correct answer is 1050.
