Question

Let $a_1,a_2,a_3....,a_{49}$ 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. $34$
• B. $33$
• C. $66$
• D. $68$

Answer 1

The correct option is A $34$

$\Rightarrow n^{th}$ term $a_n=a+(n-1)d$

$\Rightarrow$$a_9+a_{43}=66$

$\therefore a+8d+a+42d=66$

$\therefore a+25d=33$ ... (1)

Now, ${\sum }_{k=0}^{12}{a}_{4k+1}=416$

$\therefore 13a+312d=416$ (using sum of AP on the common difference parts)

$\therefore a+24d=32$ ... (2)

from (1) and (2), we get $d=1$ and $a=8$

$\therefore {\sum }_{k=1}^{17}{a}_k^2=8^2+9^2\cdots +{24}^2$

$=(1^2+2^2\cdots +{24}^2)-(1^2+2^2\cdots +7^2)$

$=\frac{24\times 25\times 49}{6}-\frac{7\times 8\times 15}{6}$ (using sum of squares of $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$

$=4760$

$=140\times 34$

Hence, the required answer is $34$.