The regular numbers lying somewhere in the range of \[100\text{ }and\text{ }1000,\]which are products of \[5,\text{ }are\text{ }105,\text{ }110,\text{ }\ldots \text{ }995.\]

It plainly frames an arrangement in A.P.

Where, the initial term, \[a\text{ }=\text{ }105\]

Normal contrast, \[d\text{ }=\text{ }5\]

Presently,

\[\begin{array}{*{35}{l}}

a\text{ }+\text{ }\left( n\text{ }-1 \right)d\text{ }=\text{ }995  \\

105\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\left( 5 \right)\text{ }=\text{ }995  \\

105\text{ }+\text{ }5n\text{ }\text{ }5\text{ }=\text{ }995  \\

5n\text{ }=\text{ }995\text{ }\text{ }105\text{ }+\text{ }5\text{ }=\text{ }895  \\

n\text{ }=\text{ }895/5  \\

n\text{ }=\text{ }179  \\

\end{array}\]

We know,

\[{{S}_{n}}~=\text{ }n/2\text{ }\left[ 2a\text{ }+\text{ }\left( n-1 \right)d \right]\]

In this manner, the amount of all regular numbers lying somewhere in the range of \[100\text{ }and\text{ }1000,\]which are products of \[5,\text{ }is\text{ }98450\].