Sequence And Series Question 86
Question: If $ a_1,\ a_2,,a_3,……a_{24} $ are in arithmetic progression and $ a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225 $ , then $ a_1+a_2+a_3+……..+a_{23}+a_{24}= $
[MP PET 1999; AMU 1997]
Options:
A) 909
B) 75
C) 750
D) 900
Show Answer
Answer:
Correct Answer: D
Solution:
$ a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225 $
$ \Rightarrow $ $ (a_1+a_{24})+(a_5+a_{20})+(a_{10}+a_{15})=225 $
$ \Rightarrow $ $ 3(a_1+a_{24})=225 $
$ \Rightarrow $ $ a_1+a_{24}=75 $ ( $ \because $ In an A.P. the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and last term) $ a_1+a_2+……+a_{24}=\frac{24}{2}(a_1+a_{24})=12\times 75=900 $ .
BETA
