Question

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

• A. $34$
• B. $33$
• C. $66$
• D. $68$

Answer 1

The correct option is A

$34$

Determine the value fo $\text{m}$

An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant.

$a_n=a_1+\left(n-1\right)d$ where $a_1=\text{first term},d=\text{common difference}$

Step 1: Calculate value of $a_{24}$

$$\begin{gathered} a_9+a_{43}=66\text{[Given]} \\ \Rightarrow \left[a_1+\left(9-1\right)d\right]+\left[a_1+\left(43-1\right)d\right]=66 \\ \Rightarrow 2a_1+\left(8+42\right)d=66 \\ \Rightarrow 2a_1+50d=66 \\ \Rightarrow 2\left(a_1+25d\right)=66 \\ \Rightarrow a_1+25d=33 \\ \Rightarrow a_{24}=33\to \left(i\right) \end{gathered}$$

Step 2: Calculate value of $a_{23}$

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

$$\begin{gathered} \sum _{k=0}^{12}{a}_{4k+1}=416 \\ \Rightarrow a_1+a_5+...+a_{49}=416 \\ \Rightarrow a_1+\left[a_1+\left(5-1\right)d\right]+...+\left[a_1+\left(49-1\right)d\right]=416 \\ \Rightarrow 13a_1+\left(4+8+...+48\right)d=416 \\ \Rightarrow 13a_1+4\left(1+2+...+12\right)d=416 \\ \Rightarrow 13a_1+312d=416\left[\because 1+2+...+n=\frac{n(n+1)}{2}\right] \\ \Rightarrow 13\left(a_1+24d\right)=416 \\ \Rightarrow a_1+24d=32 \\ \Rightarrow a_{23}=32\to \left(ii\right) \end{gathered}$$

Step 3: From $\left(i\right)$ and $\left(ii\right)$ calculate $d$

$$\begin{gathered} d=a_{24}-a_{23} \\ =33-32 \\ =1 \end{gathered}$$

substitute $d$ in $\left(i\right)$

$$\begin{gathered} a_{24}=a_1+23d \\ 33=a_1+23\times 1 \\ \Rightarrow a_1=8 \end{gathered}$$

Step 4: Determine the value of $\text{m}$.

$$\begin{gathered} {a_1}^2+{a_2}^2+...{+a_{17}}^2=140\text{m [Given]} \\ {8}^2+9^2+...{+24}^2=140\text{m} \\ \Rightarrow \left(1^2+2^2+...{+24}^2\right)-\left(1^2+2^2+...{+7}^2\right)=140\text{m} \\ \Rightarrow \left[\frac{24\times 25\times 49}{6}\right]-\left[\frac{7\times 8\times 15}{6}\right]=140\text{m}\left[\because 1^2+2^2+...+n^2=\frac{n(n+1)\left(2n+1\right)}{6}\right] \\ \Rightarrow 4900-140=140\text{m} \\ \Rightarrow \text{m}=\frac{4760}{140} \\ \Rightarrow \text{m}=34 \end{gathered}$$

Hence, option (A) is the correct answer.