JEE Main On 08 April 2018 Question 16
Question: Let $ a _1,a _2,a _3,…..,a _{49} $ be in A.P. such that $ \sum\limits _{k=0}^{12}{{a _{4k+1}}}=416 $ and $ a _9+a _{43}=66. $ If $ a_1^{2}+a_2^{2}+…….+a _17^{2}=140m $ , then $ m $ is equal to: [JEE Main Online 08-04-2018]
Options:
A) 34
B) 33
C) 66
D) 68
Show Answer
Answer:
Correct Answer: A
Solution:
$ a _1+a _5+…….+a _{49}=416 $
$ a _1+a _1+4d+…….+a _1+48d=416 $
$ 13a _1+4d( \frac{12\times 13}{2} )=416 $ $ 13(a _1+24d)=416 $
$ \therefore a _1+24d=32 $ ??(1) Also, $ a _9+a _{43}=66 $
$ \therefore 2a _1+50d=66 $ $ a _1+25d=33 $ ?..(2) From (1) & (2), we get $ d=1 $ & $ a _1=8 $
$ \therefore $ $ a _1^{2}+a _2^{2}+……+a _17^{2} $ $ =8^{2}+9^{2}+……+24^{2} $ $ =\Sigma 24^{2}-\Sigma 7^{2} $ $ =140\times 34 $
