Question

Find $7+10\frac{1}{2}+14+...+84$

Answer 1

Step 1: Given:

Here, first term, $a=7$,

Common difference, $d=a_2-a_1=10\frac{1}{2}-7=\frac{21}{2}-7=\frac{7}{2}$

and final term, $a_n=84$

Step 2: Finding number of terms, $n$ :

An arithmetic progression is defined as, $a_n=a+(n-1)d$

Thus,

$$\begin{gathered} 84=7+(n-1)·\frac{7}{2} \\ \Rightarrow n-1=\frac{77\times 2}{7} \\ \Rightarrow n=23 \end{gathered}$$

Therefore, number of term, $n=23$

Step 3: Finding the sum of the series :

Sum of an arithmetic progression ($S_n$) is given by,

$$\begin{gathered} S_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{11}{2}\left[2\times 7+(23-1)·\frac{7}{2}\right] \\ =1046\frac{1}{2} \end{gathered}$$

Therefore, $7+10\frac{1}{2}+14+...+84=1046\frac{1}{2}$