Question

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

Answer 1

The correct option is B

$34$

Step 1: Simplify the summation

$\sum _{k=0}^{12}{a}_{4k+1}=a_1+a_5+a_9+....a_{49}$

$a_1+a_5+a_9+....a_{49}=a+(a+4d)+(a+8d)+....(a+48d)$ $...\left(\because a_n=a+(n-1)d\right)$

$\Rightarrow$ $a_1+a_5+a_9+....a_{49}=13a+312d=416$

$\Rightarrow$ $13a+312d=416$

$\Rightarrow$ $a+24d=32$ $...\left(i\right)$

Step 2: Obtain second equation for first term and common difference

Given, $a_9+a_{43}=66$

$a_9+a_{43}=a+8d+a+42d$

$\Rightarrow$ $2a+50d=66$

$\Rightarrow$ $a+25d=33$ $...\left(ii\right)$

Step 3: Solve the results obtained

Subtracting $\left(i\right)$ from $\left(ii\right)$ we get

$d=1$

Re-substituting $d=1$ in $\left(i\right)$ we get

$a=8$

Step 4 : Solve for $m$

${a_1}^2+{a_2}^2+.......+{a_{17}}^2=8^2+9^2+.....{24}^2$

$\Rightarrow$${a_1}^2+{a_2}^2+.......+{a_{17}}^2=(1^2+2^2+3^2+...+{24}^2)-(1^2+2^2+3^2+...+7^2)$

Using the formula of sum of squares of first $n$ natural numbers we get

${a_1}^2+{a_2}^2+.......+{a_{17}}^2=\frac{24\times 25\times 49}{6}-\frac{7\times 8\times 15}{6}$ $....[S_{n^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}]$

$\Rightarrow$ ${a_1}^2+{a_2}^2+.......+{a_{17}}^2=4760=140m$

$\Rightarrow$ $m=\frac{4760}{140}=34$

Hence option $\left(B\right)$ is the correct answer.