Question

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

Answer 1

Given, AP $20,19\frac{1}{3},18\frac{2}{3},...$

or AP $20,\frac{58}{3},\frac{56}{3},...$

where, a = 20
and the difference $\left(d\right)=\frac{58}{3}-20=\frac{58-60}{3}=\frac{-2}{3}$
also given that $S_n$ = 300

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\frac{n}{2}[2\times 20+(n-1\left)\frac{-2}{3}\right]=300n[40-\frac{2(n-1)}{3}]=600n\left[\frac{120-2n+2}{3}\right]=600122n-2n^2=1800=122n-2n^2-1800=0n^2-61+900=0(n-36)(n-25)=0n=36or25$

Verification:

For n = 25

$S_{25}=\frac{25}{2}[2\times 20+(25-1\left)\frac{-2}{3}\right]=\frac{25}{2}[40-24\times \frac{2}{3}]=\frac{25}{2}[40-16]=\frac{25}{2}\times 24=300$

it satisfies for n = 25.

$S_{36}=\frac{36}{2}[2\times 20+(36-1\left)\frac{-2}{3}\right]=18[40-35\times \frac{2}{3}]=18[40-\frac{70}{3}]=18\times \left(\frac{120-70}{3}\right)=6\times 50=300$

it satisfies for n = 36.

So let's find the sum of terms from 26th to 36th term(the 11 terms)

Let first term be 26th term and last term be 36th term.

$S_{26to36}=\frac{11}{2}[a_{26}+a_{36}]$

$a_{26}=20+(26-1)\frac{-2}{3}=20-25\times \frac{2}{3}=20+\frac{50}{3}=\frac{60-50}{3}=\frac{10}{3}$

Similarly, $a_{36}=20+(36-1)\frac{-2}{3}=20-35\times \frac{2}{3}=20+\frac{70}{3}=\frac{60-70}{3}=\frac{-10}{3}$

therefore,

$S_{26to36}=\frac{11}{2}[a_{26}+a_{36}]=\frac{11}{2}[\frac{10}{3}+\frac{-10}{3}]=0$

So we can see that, $S_{36}=S_{25}+S_{26to36}=S_{25}+0=S_{25}$