Question

How many terms of the series (-4)+(-1)+2+5….yield a sum of 1022?

Answer 1

Given:

The series is,

`-4-1+2+5+\text{.}\ldots.\ldots\ldots`

The common difference is 3.

The sum of first n terms of the given arithmetic series is given as,

`\begin{gathered} S_n=(n)/(2)\lbrack2a+(n-1)d\rbrack \\ a=\text{ first term}=-4 \\ d=\text{ common difference}=3 \\ S_n=1022 \\ 1022=(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=(-\left(-11\right)\pm√(\left(-11\right)^2-4\cdot\:3\left(-2044\right)))/(2\cdot\:3) \\ n=(11\pm\: 157)/(6) \\ n=(11+157)/(6),n=(11-157)/(6) \\ n=(168)/(6),n=-(146)/(6) \\ n=28,n=-(73)/(3) \\ \text{But n=-}(73)/(3)\text{ is not possible value} \end{gathered}`

It gives n=28.

Answer: the 28 terms of the given series yield a sum of 1022.