Chapter 11.3, Problem 47PPSE

Solution

To identify:How many terms are in the series. The first term of an arithmetic series is 123. The common difference is 12 and the sum 1320.

The number of terms in arithmetic series is $n=8$

Given information:

The given is

  $\begin{array}{l}{a}_1=123\\ d=12\\ S=1320\end{array}$

Explanation:

Consider the information.

Arithmetic series sum formula is

  $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\mathrm{...}\left(1\right)$

Substituting all values in equation (1), and solving as

  $\begin{array}{l}{S}_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\mathrm{...}\left(1\right)\\ 1320=\frac{n}{2}\left[2\cdot 123+\left(n-1\right)12\right]\\ 1320=\frac{2n}{2}\left[123+\left(n-1\right)6\right]\\ 1320=n\left[123+6n-6\right]\\ 1320=n\left[117+6n\right]\\ 1320=117n+6n^2\\ 6n^2+117n-1320=0\end{array}$

This is a quadratic equation of the form.

This can be evaluated using quadratic formula is,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=6,b=117,andc=-1320.$

Substituting values in quadratic formula and solving as

  $\begin{array}{l}n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ n=\frac{-117\pm \sqrt{{\left(117\right)}^2-4\cdot 6\left(-1320\right)}}{2\cdot 6}\\ n=\frac{-117\pm \sqrt{45369}}{12}\\ n=\frac{-117\pm 213}{12}\end{array}$

Since numbers of terms n cannot be negative so consider only positive value as

  $\begin{array}{l}n=\frac{-117+213}{12}\\ n=\frac{96}{12}\\ n=8\end{array}$

Hence number of terms in arithmetic series is $n=8$