Question

The number of terms in the sequence for which the sum of the terms of A.P. $20,19\frac{1}{3},18\frac{2}{3},...$ is 300 is/are

• A. 25
• B. 26
• C. 31
• D. 36

Answer 1

Answer: (A), (D)

The correct options are
A 25
D 36

The given A.P. $20,19\frac{1}{3},18\frac{2}{3},...$has first term a=20 and common difference
d= $19\frac{1}{3}-20=-\frac{2}{3}$.
$S_n=\frac{n}{2}\left(2a+(n-1)d\right)\Rightarrow 300=\frac{n}{2}\left(2\times 20+(n-1)\left(-\frac{2}{3}\right)\right)\Rightarrow 600=n\left(40-\frac{2n}{3}+\frac{2}{3}\right)\Rightarrow 600=n\left(\frac{120-2n+2}{3}\right)\Rightarrow 1800=n\left(122-2n\right)\Rightarrow 1800=2n(61-n)\Rightarrow 900=61n-n^2\Rightarrow n^2-61n+900=0\Rightarrow \left(n-25\right)\left(n-36\right)=0\Rightarrow n=25or36.$
Sum of 25 terms is equal sum of 36 terms, which is 300.
This is because the common difference is negative, therefore terms go on diminishing and at one point become 0 and all terms after that are negative. Hence the sum of 26th term to 36th terms is zero and hence again we get the final sum 300.