The sum of the first 20 terms common between the series 3 + 7 + 11 + 15 + ... and 1 + 6 + 11 + 16 + ... is:

We have two arithmetic progressions (A.P.s):

First A.P. : $$3,\,7,\,11,\,15,\ldots$$

Second A.P. : $$1,\,6,\,11,\,16,\ldots$$

For any A.P. the general (n-th) term is given by the well-known formula

$$T_n = a + (n-1)d,$$

where $$a$$ is the first term and $$d$$ the common difference.

Applying this formula to the first A.P. we get

$$T_n^{(1)} = 3 + (n-1)\,4.$$

Similarly, for the second A.P. we have

$$T_m^{(2)} = 1 + (m-1)\,5.$$

Now we look for the common terms, that is, all numbers which are simultaneously equal to some $$T_n^{(1)}$$ and some $$T_m^{(2)}$$. Hence we set

$$3 + (n-1)\,4 = 1 + (m-1)\,5.$$

Expanding both sides gives

$$3 + 4n - 4 \;=\; 1 + 5m - 5.$$

Simplifying each side separately,

$$4n - 1 \;=\; 5m - 4.$$

Rearranging terms to group the variables,

$$4n \;=\; 5m - 3.$$

Or, equivalently,

$$5m = 4n + 3.$$

Because the left side $$5m$$ is a multiple of $$5$$, the right side $$4n + 3$$ must also be a multiple of $$5$$. Therefore we impose the divisibility condition

$$4n + 3 \equiv 0 \pmod{5}.$$

Noting that $$4 \equiv -1 \pmod{5},$$ we rewrite the congruence:

$$-n + 3 \equiv 0 \pmod{5}.$$

Which is the same as

$$-n \equiv -3 \pmod{5},$$

and hence

$$n \equiv 3 \pmod{5}.$$

Thus $$n$$ can be expressed as

$$n = 5k + 3,$$

where $$k$$ is any non-negative integer $$k = 0,1,2,\ldots$$.

Substituting this value of $$n$$ back into the expression for the term of the first A.P. we obtain every common term:

$$\begin{aligned} T &= 3 + \bigl[(5k + 3) - 1\bigr]\;4 \\ &= 3 + (5k + 2)\,4 \\ &= 3 + 20k + 8 \\ &= 20k + 11. \end{aligned}$$

So the sequence of all common terms is itself an A.P.:

$$11,\,31,\,51,\,71,\ldots$$

From the form $$20k + 11$$ we see the first common term is $$11$$ and the common difference is $$20$$.

We are asked for the sum of the first $$20$$ such common terms. Hence we take

$$a = 11,\quad d = 20,\quad n = 20.$$

For an A.P. the sum of the first $$n$$ terms is given by the formula

$$S_n = \frac{n}{2}\Bigl[\,2a + (n-1)d\,\Bigr].$$

Substituting $$n = 20,\; a = 11,\; d = 20,$$ we have

$$\begin{aligned} S_{20} &= \frac{20}{2}\Bigl[\,2(11) + (20-1)(20)\Bigr] \\ &= 10\Bigl[\,22 + 19 \times 20\Bigr] \\ &= 10\Bigl[\,22 + 380\Bigr] \\ &= 10 \times 402 \\ &= 4020. \end{aligned}$$

Hence, the correct answer is Option B.