Question

How many terms of the ap 20 19 1/3 ,18⅔............Be taken so that their sum is 300?

Answer 1

given  20,19 1/3 ,18⅔............ are in AP

and $s_n =300$

a=20 and d=$19_{1/2}$-20= -2/3

$s_n=\frac{n}{2} *(2a+(n-1)*d)$... formula

then

300=$\frac{n}{2} *(2*20+(n-1)*-2/3$

600=n$\frac{120-2n+2}{3}$

1800=120n-2n^2+2n

900=61n-n^2

n^2 - 61n+900

n=36 or n=26

hope it helps...