Question

In Problems 37–54, find each sum.

$\sum _{n=1}^{100}(6-\frac{1}{2}n)$

Short answer

$\sum _{n=1}^{100}(6-\frac{1}{2}n)=-1925$

Step-by-step solution

Given Information 

$\sum _{n=1}^{100}(6-\frac{1}{2}n)$

first term 

use the given sum to find out the first term a1

$$\begin{gathered} \sum _{n=1}^{100}(6-\frac{1}{2}n) \\ {a}_n=(6-\frac{1}{2}n) \\ Tofind{a}_1,substituten=1 \\ {a}_1=(6-\frac{1}{2}(1\left)\right)=\frac{11}{2} \end{gathered}$$

find out last term 

To find out last term , we substitute n=100

$$\begin{gathered} a_n=(6-\frac{1}{2}n) \\ {a}_{100}=(6-\frac{1}{2}(100\left)\right)=-44 \end{gathered}$$

Sum 

To find out sum we use sum formula

$$\begin{gathered} S_n=\frac{n}{2}(a_1+a_{100}) \\ Sumof100terms \\ {S}_{100}=\frac{100}{2}(\frac{11}{2}+-44) \\ {S}_{100}=50\left(\frac{11}{2}-\frac{88}{2}\right) \\ {S}_{100}=25\left(-77\right) \\ {S}_{100}=-1925 \end{gathered}$$