Question

Find the number of terms in the series $20+19\frac{1}{3}+18\frac{2}{3}+....$ of which the sum is $300$, explain the double answer.

Answer 1

The given series is an arithmetic series with first term $a(=20)$ and the common difference $d(=-\frac{2}{3})$.

Let the sum of $n$ terms be $300$.

Then, $S_n=300$

$⟹\frac{n}{2}\{2a+(n-1\left)d\right\}=300$
$⟹\frac{n}{2}\{2\times 20+(n-1\left)\right(-2/3\left)\right\}=300$
$⟹n^2-61n+900=0⟹(n-25)(n-36)=0⟹n=25$ or $36$
So, sum of $25$ terms $=$ sum of $36$ terms $=300$

Here, the common difference is negative therefore terms go on diminishing and ${31}^{st}$ term becomes zero.

All terms after ${31}^{st}$ term are negative. These negative terms when added to positive terms from ${26}^{th}$ term to ${30}^{th}$ term, they cancel out each other and the sum remains the same.

Hence, the sum of $25$ terms as well as that of $36$ terms is $300$.