How many terms of the AP

Question:

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

 

Solution:

To Find: Number of terms required to make the sum of the AP 300.

Let the first term of the AP be a and the common difference be d

Here $\mathrm{a}=20, d=-\frac{2}{3}$

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

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

$\Rightarrow 300 \times 6=n[120-2(n-1)]$

$\Rightarrow n[-2 n+122]=6 \times 300$

$\Rightarrow n(-n+61)=3 \times 300$

$\Rightarrow n=36$ or 25

Explanation: Since the given AP is a decreasing progression where $a_{n-1}>a_{n}$, it is bound to have negative values in the series. $S_{n}$ is maximum for $n=30$ or $n=31\left(S_{30}=S_{31}=\right.$ $\left.S_{\max }=310\right) .$ The sum of 300 can be attained by either adding 25 terms or 36 terms so that negative terms decrease the maximum sum to 300 .

 

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