Question

How many terms of the AP ${\displaystyle 20,19\frac{1}{3},18\frac{2}{3},...}$ must be taken so that their sum is 300? Explain the double answer.

Answer 1

The given AP is: ${\displaystyle 20,19\frac{1}{3},18\frac{2}{3},...}$

First term (a): a = 20

Common difference (d): d = ${\displaystyle 19\frac{1}{3}-20=-\frac{2}{3}}$

The sum of the first n terms of an AP is given by:

${\displaystyle S_n=300,a=20,\text{and}d=-\frac{2}{3}}$, we get 

${\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]}$

Substituting S_n = 300, a = 20, and d = ${\displaystyle -\frac{2}{3}}$, we get:

${\displaystyle 300=\frac{n}{2}[2\left(20\right)+(n-1)(-\frac{2}{3})]}$

${\displaystyle 300=\frac{n}{2}[40-\frac{2}{3}(n-1)]}$

Multiply through by 2 to eliminate the fraction:

${\displaystyle 600=n[40-\frac{2}{3}(n-1)]}$

${\displaystyle 600=n[40-2\frac{n}{3}+\frac{2}{3}]}$

${\displaystyle 600=n[\frac{120}{3}-\frac{2n}{3}+\frac{2}{3}]}$

${\displaystyle 600=n\left[\frac{122-2n}{3}\right]}$

1800 = n(122 − 2n)

1800 = 122n − 2n²

2n² − 122n + 1800 = 0

Divide through by 2 to simplify: n² − 61n + 900 = 0

${\displaystyle n=\frac{-b\pm {\sqrt{b}}^2-4ac}{2a}}$

${\displaystyle n=\frac{-(-61)\pm \sqrt{{(-61)}^2-4\left(1\right)\left(900\right)}}{2}}$

${\displaystyle n=\frac{61\pm \sqrt{3721-3600}}{2}}$

${\displaystyle n=\frac{61\pm \sqrt{121}}{2}}$

${\displaystyle n=\frac{61\pm 11}{2}}$

${\displaystyle n=\frac{61+11}{2}=\frac{72}{2}=36}$

${\displaystyle n=\frac{61-11}{2}=\frac{50}{2}=25}$

The two solutions, n = 25 and n = 36, occur because the AP has a negative common difference (d = ${\displaystyle -\frac{2}{3}}$​), meaning the terms decrease as n increases.