Question

How many terms of sequence
$20+19\frac{1}{3}+18\frac{2}{3}+...$ must be taken so that their sum becomes $300$. Also explain the reason of double answers.

Answer 1

$20+\frac{58}{3}+\frac{56}{3}+...$

$=\frac{60}{3}+\frac{58}{3}+\frac{56}{3}+...$

This forms an AP with

$a=60/3d=\frac{-2}{3}$

$Sn=\frac{n}{2}[2a+(n-1)d]=300$

$\Rightarrow \frac{n}{2}[40+(n-1)\left(-2/3\right)]=300$

$\Rightarrow 20n-\frac{(n-1)n}{3}=300$

$\Rightarrow 60n-n^2+n=900$

$\Rightarrow n^2-61n+900=0$

$n^2-25n-36n+900=0$

$n(n-36)-25(n-36)=0$

$n=25,3.6$

First $05$ terms gives the sum of $300$

In the next $II$ terms, $1$ term is $0$

and term $26$ to $30$ is negative of terms $32$ to $36$.So we get $2$ values of n.