The common difference of the A.P. $$b_1, b_2, \ldots, b_m$$ is 2 more than common difference of A.P. $$a_1, a_2, \ldots, a_n$$. If $$a_{40} = -159$$, $$a_{100} = -399$$ and $$b_{100} = a_{70}$$, then $$b_1$$ is equal to:

First, recall the $$n$$-th term formula for an arithmetic progression (A.P.):

$$\text{For an A.P. } x_1, x_2,\ldots :\; x_n = x_1 + (n-1)d,$$

where $$d$$ is the common difference.

For the A.P. $$a_1, a_2, \ldots,$$ let the first term be $$a_1$$ and common difference be $$d_a$$. We are given

$$a_{40} = -159, \qquad a_{100} = -399.$$

Applying the formula to $$a_{40}$$, we have

$$a_{40} = a_1 + (40-1)d_a = a_1 + 39d_a = -159 \quad\text{…(1)}$$

Similarly, for $$a_{100}$$,

$$a_{100} = a_1 + (100-1)d_a = a_1 + 99d_a = -399 \quad\text{…(2)}$$

Now, subtract equation (1) from equation (2):

$$\bigl(a_1 + 99d_a\bigr) - \bigl(a_1 + 39d_a\bigr) = -399 - (-159).$$

This simplifies to

$$60d_a = -240,$$

so

$$d_a = \frac{-240}{60} = -4.$$

Substituting $$d_a = -4$$ back into equation (1),

$$a_1 + 39(-4) = -159 \;\;\Longrightarrow\;\; a_1 - 156 = -159,$$

hence

$$a_1 = -159 + 156 = -3.$$

Thus, for the $$a$$-sequence we have $$a_1 = -3$$ and $$d_a = -4.$$

Next, consider the A.P. $$b_1, b_2, \ldots$$ with first term $$b_1$$ and common difference $$d_b$$. The problem states that this common difference is “2 more” than that of the $$a$$-sequence, i.e.

$$d_b = d_a + 2 = -4 + 2 = -2.$$

We are also told that $$b_{100} = a_{70}.$$ First evaluate $$a_{70}$$ using the $$a$$-sequence formula:

$$a_{70} = a_1 + (70-1)d_a = -3 + 69(-4).$$

Calculating gives

$$a_{70} = -3 - 276 = -279.$$

Therefore,

$$b_{100} = -279.$$

Now apply the term formula to the $$b$$-sequence for $$n = 100$$:

$$b_{100} = b_1 + (100-1)d_b = b_1 + 99d_b.$$

Substituting $$d_b = -2$$ and $$b_{100} = -279$$ gives

$$b_1 + 99(-2) = -279.$$

Simplifying,

$$b_1 - 198 = -279,$$

so

$$b_1 = -279 + 198 = -81.$$

Hence, the correct answer is Option C.