Question

Let $a_1, a_2, a_3, ………, a_{49}$ be in A.P. such that $\displaystyle\sum^{12}_{k = 0} a_{4k + 1 } = 416$ and $a_9 + a_{43} = 66 $. If $a_1^2 + a_2^2 + .... + a_{17}^2 = 140 \, m $, 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, $a_{1}+a_{5}+a_{9}+\ldots \ldots+a_{49}=416$
$\Rightarrow a+24 d=32\,\,\,\,\,\,\,\,...(i)$
Also, $a_{9}+a_{43}=66 $
$\Rightarrow a+25 d=33\,\,\,\,\,\,\,\,\,...(ii)$
Solving (i) & (ii),
We get $d=1, a=8$
Now, $a_{1}^{2}+a_{2}^{2}+\ldots . .+a_{17}^{2}=140\, m$
$\Rightarrow 8^{2}+9^{2}+\ldots . .+24^{2}=140\, m$
$\Rightarrow \frac{24 \times 25 \times 49}{6}-\frac{7 \times 8 \times 15}{6}=140\, m$
$\Rightarrow m=34$