Counting by twos, write numbers starting from:
(a) 7786
(b) 4003.
Counting by fives, write numbers starting from:
(a) 3505
(b) 7913.

Hint: We express the increment of the given number by a fixed digit can be expressed by adding that number to the previous number. The addition is only applicable to the previous number to get the following number.

Complete step by step solution:
We start counting by a fixed number by adding that number to the previous number.
Therefore, to get the count by twos starting from 7786, we need to keep adding 2.
So, the next numbers will be
$\begin{align}
  & 7786+2=7788 \\
 & 7788+2=7790 \\
 & 7790+2=7792 \\
 & ............. \\
\end{align}$
Similarly, to get the count by twos starting from 4003, we need to keep adding 2.
So, the next numbers will be
$\begin{align}
  & 4003+2=4005 \\
 & 4005+2=4007 \\
 & 4007+2=4009 \\
 & ............. \\
\end{align}$
Similarly, to get the count by fives starting from 3505, we need to keep adding 5.
So, the next numbers will be
$\begin{align}
  & 3505+5=3510 \\
 & 3510+5=3515 \\
 & 3515+5=3520 \\
 & ............. \\
\end{align}$
Similarly, to get the count by fives starting from 7913, we need to keep adding 5.
So, the next numbers will be
$\begin{align}
  & 7913+5=7918 \\
 & 7918+5=7923 \\
 & 7923+5=7928 \\
 & ............. \\
\end{align}$

Note: The new series of the numbers become an A.P. As the difference between the numbers is fixed the progression remains in series form. We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term be ${{t}_{1}}$ and the common difference be $d$ where $d={{t}_{2}}-{{t}_{1}}={{t}_{3}}-{{t}_{2}}={{t}_{4}}-{{t}_{3}}$.
We can express the general term ${{t}_{n}}$ based on the first term and the common difference.
The formula being ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d$.