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

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 seriesStep 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]