Respuesta :
The sum of first 20 arithmetic series [tex]S_{20}=\frac{-3475}{16}[/tex]
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 [tex]a_{3}=a_{1}+(3-1) d[/tex]
[tex]a_{1}+2 d=55[/tex]
Arithmetic series for 19th term is
[tex]a_{19}=a_{1}+(19-1) d=-98[/tex]
[tex]a_{19}+18 d=-98[/tex]
Subtracting equation 2 from 1
[tex]\left[a_{19}+18 d=-98\right]+\left[a_{1}+2 d=55\right][/tex]
16d=-98-55
16d=-153
[tex]d=\frac{-153}{16}[/tex]
Also we know[tex]a_{1}+2 d=55[/tex]
[tex]a_{1}+2(-153 / 16)=55[/tex]
[tex]a_{1}+(-153 / 8)=55[/tex]
[tex]a_{1}=55+(153 / 8)[/tex]
[tex]a_{1}=440+153 / 8[/tex]
[tex]a_{1}=553 / 8[/tex]
First 20 terms of an AP
[tex]a_{n=} a_{1}+(n-1) d[/tex]
[tex]a_{20}=553 / 8+19(-153 / 16)[/tex]
[tex]a_{20}=553 / 8+19(-153 / 16)[/tex]
[tex]a_{20}=\{553 * 2 / 8 * 2\}-2907 / 16[/tex]
[tex]a_{20}=[1106 / 16]-[2907 / 16][/tex]
[tex]a_{20}=-1801 / 16[/tex]
Sum of 20 Arithmetic series is
[tex]S_{n}=n\left(a_{1}+a_{n}\right) / 2[/tex]
Substitute the known values in the above equation we get
[tex]S_{20}=\left[\frac{20\left(\left(\frac{558}{8}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right][/tex]
[tex]S_{20}=\left[\frac{\left.20\left(\frac{1106}{16}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right][/tex]
[tex]S_{20}=10 \frac{(-695 / 16)}{2}[/tex]
[tex]S_{20}=5\left[\frac{-695}{16}\right][/tex]
[tex]S_{20}=\frac{-3475}{16}[/tex]
Result:
Thus the sum of first 20 terms in an arithmetic series is [tex]S_{20}=\frac{-3475}{16}[/tex]
