Sequence And Series Question 482
Question: The sum of the numbers between 100 and 1000 which is divisible by 9 will be
[MP PET 1982]
Options:
A) 55350
B) 57228
C) 97015
D) 62140
Show Answer
Answer:
Correct Answer: A
Solution:
Series $ 108+117+……..+999 $ is an A.P. where $ a=108 $ , common difference $ d=9 $ , $ n=\frac{999}{9}-\frac{99}{9}=111-11=100 $ Hence required sum = $ \frac{100}{2}(108+999)=50\times 1107=55350 $ .
BETA
