If the sum of the first 40 terms of the series, $$3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + \ldots$$ is $$(102)m$$, then $$m$$ is equal to

We begin by writing a few terms again to notice the regularity:

$$3,\;4,\;8,\;9,\;13,\;14,\;18,\;19,\;\ldots$$

Inside each small block of two successive terms the difference is $$1$$, while the first number of one block is obtained from the first number of the previous block by adding $$5$$. The same happens with the second numbers. Hence every block (or “pair”) can be described neatly.

Let us label the blocks by a non-negative integer $$k$$.

For $$k=0$$ the two terms are $$3$$ and $$4$$.

Adding $$5$$ once gives the next block $$8,9$$, so for $$k=1$$ the terms are $$3+5(1)=8$$ and $$4+5(1)=9$$.

Continuing like this, the block numbered $$k$$ contributes the two terms

$$3+5k \qquad\text{and}\qquad 4+5k.$$

Hence the entire series may be written as

$$\bigl(3+5\!\cdot\!0,\;4+5\!\cdot\!0\bigr),\; \bigl(3+5\!\cdot\!1,\;4+5\!\cdot\!1\bigr),\; \bigl(3+5\!\cdot\!2,\;4+5\!\cdot\!2\bigr),\;\ldots$$

In every block there are two terms, so the first $$40$$ terms form $$40/2=20$$ complete blocks. Thus we must sum the blocks for $$k=0,1,2,\ldots,19$$.

Within a single block the sum of the two terms is

$$\bigl(3+5k\bigr)+\bigl(4+5k\bigr)=7+10k.$$

Therefore the required sum of the first $$40$$ terms is

$$S_{40}=\sum_{k=0}^{19}\bigl(7+10k\bigr).$$

We separate the constant part and the part involving $$k$$:

$$S_{40}= \sum_{k=0}^{19}7 \;+\; \sum_{k=0}^{19}10k = 7\sum_{k=0}^{19}1 \;+\;10\sum_{k=0}^{19}k.$$

The first summation is simply $$7$$ added $$20$$ times, giving $$7\times20$$.

For the second summation we use the standard formula for the sum of the first $$n$$ natural numbers:

$$\sum_{k=0}^{n-1}k = \frac{n(n-1)}{2}.$$

Here $$n=20$$, so

$$\sum_{k=0}^{19}k = \frac{20(20-1)}{2}=\frac{20\cdot19}{2}=190.$$

Substituting these results back, we get

$$S_{40}=7\times20 \;+\;10\times190 = 140 \;+\;1900 = 2040.$$

The problem states that this sum equals $$(102)m$$. Hence

$$2040 = 102\,m.$$

Solving for $$m$$ we divide both sides by $$102$$:

$$m=\frac{2040}{102}=20.$$

Hence, the correct answer is Option A.