Question

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

• A. only 25 terms
• B. only 36 terms
• C. only 49 terms
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A only 25 terms
B only 36 terms

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

$19\frac{1}{3}-20$

$=\frac{58}{3}-20=\frac{58-60}{3}=\frac{-2}{3}$

again

$18\frac{2}{3}-19\frac{1}{3}$

$=\frac{56}{3}-\frac{58}{3}=\frac{-2}{3}$

$19\frac{1}{3}-20=18\frac{2}{3}-19\frac{1}{3}=\frac{-2}{3}=d$

$\therefore$ the given series is a A.P where first term $a=20$

$\therefore a=20,\frac{-2}{3}$

$\therefore$ Let the read.no. pf terms be ${}^{\prime }{n}^{\prime }$

$S_n=300$

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

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

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

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

$\Rightarrow [60-n+1]=900$

$\Rightarrow n[61-n]=900$

$\Rightarrow n^2-61n+900=0$

$\Rightarrow n^2-36n-25n+900=0$

$\Rightarrow n(n-36)-25(n-36)=0$

$\therefore n=25$ or $n=36$

both option $A$ and $B$.