Respuesta :
Given:
The series is,
[tex]-4-1+2+5+\text{.}\ldots.\ldots\ldots[/tex]The common difference is 3.
The sum of first n terms of the given arithmetic series is given as,
[tex]\begin{gathered} S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack \\ a=\text{ first term}=-4 \\ d=\text{ common difference}=3 \\ S_n=1022 \\ 1022=\frac{n}{2}\lbrack2(-4)+(n-1)3\rbrack \\ 2\cdot1022=n(-8+3n-3) \\ 2044=3n^2-11n \\ 3n^2-11n-2044=0 \\ \text{Use the quadratic formula,} \\ n=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a},a=3,b=-11,c=-2044 \\ n=\frac{-\left(-11\right)\pm\sqrt{\left(-11\right)^2-4\cdot\:3\left(-2044\right)}}{2\cdot\:3} \\ n=\frac{11\pm\: 157}{6} \\ n=\frac{11+157}{6},n=\frac{11-157}{6} \\ n=\frac{168}{6},n=-\frac{146}{6} \\ n=28,n=-\frac{73}{3} \\ \text{But n=-}\frac{73}{3}\text{ is not possible value} \end{gathered}[/tex]It gives n=28.
Answer: the 28 terms of the given series yield a sum of 1022.
