The Common Progression Problem

Relevant wiki: Arithmetic Progression Sum

By writing out more terms for each of these sequences of numbers, we noticed that the numbers that are in common are

$21, 41, 61, \ldots$

Since the terms that are in both sequences have a difference of 20, we get a new sequence with the direct formula: $u_n = 20n + 21$ .

Using the formula for the arithmetic progression, we have the sum of the first $n$ terms of an arithmetic progression, $S_n$ ,

$S_n = \dfrac n2 [ 2a + (n-1)d ] \; ,$

where $a=21$ is the first term, $d=20$ is the common difference. $n =100$ . Substituting these numbers gives the desired answer of

$S_{100} = \dfrac{100}2 [ 21 + (100-1)(22) ] = \boxed{101,100} \; .$

The terms common to them are $21, 41, 61, 81 ...$ . Notice that it forms an Arithmetic Progression with a common difference of $20$ . Finding the sum of the first $100$ common to them, we have

$\large s = \frac{n}{2}[2a_1 + (n - 1)d]$ $\large= 50[42 + (99)(20)]=$ $\boxed{\large\color{#624F41}101100}$