Let $$a_1, a_2, a_3, \ldots, a_{49}$$ be in A.P. such that $$\sum_{k=0}^{12} a_{4k+1} = 416$$ and $$a_9 + a_{43} = 66$$. If $$a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m$$, then m is equal to:

Let the first term of the given A.P. be $$a_1 = A$$ and let the common difference be $$d$$. For every natural number $$n$$, the general term is

$$a_n = A + (n-1)d.$$

First we translate the information $$\displaystyle\sum_{k=0}^{12} a_{4k+1}=416$$. The indices are $$1,5,9,\ldots ,49$$; there are $$13$$ of them. These terms themselves form an A.P. whose first term is $$a_1=A$$ and whose common difference is

$$a_{5}-a_{1}=4d.$$

Using the sum formula for an A.P.,

$$\frac{n}{2}\,[2(\text{first})+(n-1)(\text{common difference})] =\frac{13}{2}\,[2A+12\cdot4d]=416.$$

Simplifying,

$$\frac{13}{2}[2A+48d]=416\;\Longrightarrow\;13[2A+48d]=832 \;\Longrightarrow\;2A+48d=64.$$

Dividing by $$2$$ gives the first relation

$$A+24d=32\qquad\text{(I)}.$$

Next we use $$a_9+a_{43}=66$$. Re-express each term with the general formula:

$$a_9=A+8d,\qquad a_{43}=A+42d.$$

Therefore

$$a_9+a_{43}=2A+50d=66.$$

Dividing by $$2$$ gives the second relation

$$A+25d=33\qquad\text{(II)}.$$

Subtract (I) from (II):

$$(A+25d)-(A+24d)=33-32\;\Longrightarrow\;d=1.$$

Substituting $$d=1$$ into (I) yields

$$A+24(1)=32\;\Longrightarrow\;A=8.$$

Thus $$a_1=8$$ and $$d=1$$. We now evaluate $$\displaystyle S=\sum_{n=1}^{17} a_n^2$$.

Write $$a_n=A+(n-1)d=8+(n-1).$$ Letting $$n-1=k$$, where $$k$$ runs from $$0$$ to $$16$$, we have

$$S=\sum_{k=0}^{16} (8+k)^2 =\sum_{k=0}^{16}\!\Bigl(64+16k+k^2\Bigr).$$

Separate the sums term by term:

$$S=64\sum_{k=0}^{16}1+16\sum_{k=0}^{16}k+\sum_{k=0}^{16}k^2.$$

Use the standard results

$$\sum_{k=0}^{N}1=N+1,\quad \sum_{k=0}^{N}k=\frac{N(N+1)}{2},\quad \sum_{k=0}^{N}k^2=\frac{N(N+1)(2N+1)}{6}.$$

With $$N=16$$, these give

$$\sum_{k=0}^{16}1=17,\quad \sum_{k=0}^{16}k=\frac{16\cdot17}{2}=136,\quad \sum_{k=0}^{16}k^2=\frac{16\cdot17\cdot33}{6}=1496.$$

Substituting back,

$$S=64(17)+16(136)+1496 =1088+2176+1496 =4760.$$

The problem states $$a_1^2+a_2^2+\dots +a_{17}^2=140m$$, hence

$$140m=4760\;\Longrightarrow\;m=\frac{4760}{140}=34.$$

Hence, the correct answer is Option D.