Question

Find each sum

$8+8\frac{1}{4}+8\frac{1}{2}+8\frac{3}{4}+9+...+50$

Short answer

The solution is$S_{169}=4901$

Step-by-step solution

Given information

$8+8\frac{1}{4}+8\frac{1}{2}+8\frac{3}{4}+9+...+50$

Calculation

We have, $8+8\frac{1}{4}+8\frac{1}{2}+8\frac{3}{4}+9+...+50$

So, $a_1=8,d=\frac{1}{4},a_n=50$

Now use formula,

$$\begin{gathered} 50=8+(n-1)\frac{1}{4} \\ \Rightarrow 50=8+\frac{n}{4}-\frac{1}{4} \\ \Rightarrow \frac{n}{4}=50-8-\frac{1}{4} \\ \Rightarrow n=169 \end{gathered}$$

To find sum $S_{169}$

$$\begin{gathered} S_{169}=\frac{169}{2}(8+50) \\ =4901 \end{gathered}$$