Question

Let $a_1,a_2,a_3,\dots \dots \dots ,a_{49}$ be in A.P. such that $\sum _{k=0}^{12}{a}_{4k+1}=416$ and $a_9+a_{43}=66$. If $a_1^2+a_2^2+....+a_{17}^2=140m$, then m is equal to :

• A. 66
• B. 68
• C. 34
• D. 33

Answer 1

Answer: (C)

Let $a_1=a$ and common difference $=d$
Given, $a_1+a_5+a_9+\dots \dots +a_{49}=416$
$\Rightarrow a+24d=32...(i)$
Also, $a_9+a_{43}=66$
$\Rightarrow a+25d=33...(ii)$
Solving (i) & (ii),
We get $d=1,a=8$
Now, $a_1^2+a_2^2+\dots ..+a_{17}^2=140m$
$\Rightarrow 8^2+9^2+\dots ..+24^2=140m$
$\Rightarrow \frac{24\times 25\times 49}{6}-\frac{7\times 8\times 15}{6}=140m$
$\Rightarrow m=34$