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)=(-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=(-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[(20\left(\left((558)/(8)\right)+\left((-1801)/(16)\right)\right))/(2)\right]`

`S_(20)=\left[(\left.20\left((1106)/(16)\right)+\left((-1801)/(16)\right)\right))/(2)\right]`

`S_(20)=10 ((-695 / 16))/(2)`

`S_(20)=5\left[(-695)/(16)\right]`

`S_(20)=(-3475)/(16)`

Result:

Thus the sum of first 20 terms in an arithmetic series is
`S_(20)=(-3475)/(16)`