Question

Find the sum of the first 20 terms of an arithmetic progression of which the third term is 55 and the last term is -98

Answer 1

The sum of first 20 arithmetic series $S_{20}=\frac{-3475}{16}$

Given:

Arithmetic series for 3rd term is 55

Arithmetic series for 7th term is -98

To find:

The sum of first 20 Arithmetic series

Step by Step Explanation:

Solution:

Formula for calculating arithmetic series

Arithmetic series=a+(n-1) d

Arithmetic series for 3rd term $a_{3}=a_{1}+(3-1) d$

$a_{1}+2 d=55$

Arithmetic series for 19th term is

$a_{19}=a_{1}+(19-1) d=-98$

$a_{19}+18 d=-98$

Subtracting equation 2 from 1

$\left[a_{19}+18 d=-98\right]+\left[a_{1}+2 d=55\right]$

16d=-98-55

16d=-153

$d=\frac{-153}{16}$

Also we know$a_{1}+2 d=55$

$a_{1}+2(-153 / 16)=55$

$a_{1}+(-153 / 8)=55$

$a_{1}=55+(153 / 8)$

$a_{1}=440+153 / 8$

$a_{1}=553 / 8$

First 20 terms of an AP  

$a_{n=} a_{1}+(n-1) d$

$a_{20}=553 / 8+19(-153 / 16)$

$a_{20}=553 / 8+19(-153 / 16)$

$a_{20}=\{553 * 2 / 8 * 2\}-2907 / 16$

$a_{20}=[1106 / 16]-[2907 / 16]$

$a_{20}=-1801 / 16$

Sum of 20 Arithmetic series is

$S_{n}=n\left(a_{1}+a_{n}\right) / 2$

Substitute the known values in the above equation we get

$S_{20}=\left[\frac{20\left(\left(\frac{558}{8}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right]$

$S_{20}=\left[\frac{\left.20\left(\frac{1106}{16}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right]$

$S_{20}=10 \frac{(-695 / 16)}{2}$

$S_{20}=5\left[\frac{-695}{16}\right]$

$S_{20}=\frac{-3475}{16}$

Result:

Thus the sum of first 20 terms in an arithmetic series is $S_{20}=\frac{-3475}{16}$