1. Write the A.P. in standard form

Let the A.P. be $a_n=a+(n-1)d,$ where $a=a_1$ and $d$ is the common difference.

2. Use the first condition

We are given $\sum_{k=0}^{12} a_{4k+1}=416.$

Now, $a_{4k+1}=a+(4k)d.$ So the terms are: $a_1,a_5,a_9,\dots,a_{49}.$ This is an A.P. with 13 terms.

Hence, $\sum_{k=0}^{12} a_{4k+1}=\frac{13}{2}(a_1+a_{49})=416.$

Now, $a_{49}=a+48d.$ So, $\frac{13}{2}(a+a+48d)=416$ $\frac{13}{2}(2a+48d)=416$ $13(a+24d)=416$ $a+24d=32. \qquad (1)$

3. Use the second condition

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

Now, $a_9=a+8d, \qquad a_{43}=a+42d.$ So, $(a+8d)+(a+42d)=66$ $2a+50d=66$ $a+25d=33. \qquad (2)$

4. Solve for $a$ and $d$

Subtract (1) from (2): $(a+25d)-(a+24d)=33-32$ $d=1.$

Put into (1): $a+24=32$ $a=8.$

Thus the A.P. is $a_n=8+(n-1)=n+7.$

5. Compute $a_1^2+a_2^2+\cdots+a_{17}^2$

Since $a_n=n+7$, $a_1,a_2,\dots,a_{17}=8,9,10,\dots,24.$

Therefore, $\sum_{n=1}^{17} a_n^2=8^2+9^2+\cdots+24^2.$

This is $\sum_{r=1}^{24} r^2-\sum_{r=1}^{7} r^2.$

Using $\sum_{r=1}^{n} r^2=\frac{n(n+1)(2n+1)}{6},$ we get $\sum_{r=1}^{24} r^2=\frac{24\cdot25\cdot49}{6}=4900,$ $\sum_{r=1}^{7} r^2=\frac{7\cdot8\cdot15}{6}=140.$

So, $8^2+9^2+\cdots+24^2=4900-140=4760.$

Given $a_1^2+a_2^2+\cdots+a_{17}^2=140m,$ so $140m=4760$ $m=\frac{4760}{140}=34.$

6. Check options

The value is $m=34.$ So the correct option is D.

7. Comparison with stored answer

Stored correct answer: D

Derived answer: D

They agree.