Question

Let \(a_1,a_2,a_3,\ldots,a_{49}\) be in A.P. such that \(\sum_{k=0}^{12}a_{4k+1}=416\) and \(a_9+a_{43}=66\). If \(a_1^2+a_2^2+\cdots+a_{17}^2=140m\), then \(m\) is equal to

• A. 33
• B. 66
• C. 68
• D. 34

Hint:

Find \(a_1\) and \(d\) from the two conditions (the first sum is of 13 terms of an A.P. of common difference \(4d\)), then sum the squares.

Answer 1

The Correct Option is D

Let the A.P. have first term \(a\) and common difference \(d\).
\(a_1,a_5,\ldots,a_{49}\) are 13 terms in A.P. with sum \(\frac{13}{2}(a_1+a_{49})=416\Rightarrow a_1+a_{49}=64\Rightarrow2a+48d=64\Rightarrow a+24d=32\).
\(a_9+a_{43}=2a+50d=66\Rightarrow a+25d=33\).
Subtracting: \(d=1\), \(a=8\). So \(a_n=n+7\).
\(\sum_{n=1}^{17}(n+7)^2=\sum_{j=8}^{24}j^2=\frac{24\cdot25\cdot49}{6}-\frac{7\cdot8\cdot15}{6}=4900-140=4760\).
\(140m=4760\Rightarrow m=34\).
Answer: (4) 34