# How many terms of the AP 20,

Question:

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

Solution:

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

Here, $a=20$ and $d=19 \frac{1}{3}-20=\frac{58}{3}-20=\frac{58-60}{3}=-\frac{2}{3}$

Let the required number of terms be n. Then,

$S_{n}=300$

$\Rightarrow \frac{n}{2}\left[2 \times 20+(n-1) \times\left(-\frac{2}{3}\right)\right]=300 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

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

$\Rightarrow \frac{n}{2} \times \frac{(122-2 n)}{3}=300$

$\Rightarrow 122 n-2 n^{2}=1800$

$\Rightarrow 2 n^{2}-122 n+1800=0$

$\Rightarrow 2 n^{2}-50 n-72 n+1800=0$

$\Rightarrow 2 n(n-25)-72(n-25)=0$

$\Rightarrow(n-25)(2 n-72)=0$

$\Rightarrow n-25=0$ or $2 n-72=0$

$\Rightarrow n=25$ or $n=36$

So, the sum of first 25 terms as well as that of first 36 terms is 300. This is because the sum of all terms from 26th to 36th is 0.

Hence, the required number of terms is 25 or 36.