Sequence And Series Question 499
Question: The sum of numbers from 250 to 1000 which are divisible by 3 is
[RPET 1997]
Options:
A) 135657
B) 136557
C) 161575
D) 156375
Show Answer
Answer:
Correct Answer: D
Solution:
The number divisible by 3 between 250 to 1000 are 252, 255, ………., 999.
$ \therefore $ $ T_{n}=999=252+(n-1)3 $
$ \Rightarrow $ $ 333=84+n-1 $
$ \Rightarrow $ $ n=250 $
$ \therefore $ $ S=\frac{n}{2}[a+l]=\frac{250}{2}[252+999] $ = $ 125\times 1251=156375 $ .
BETA
