The regular numbers lying somewhere in the range of
![\[100\text{ }and\text{ }1000,\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-73c944d941928de6405ec8060ff2906b_l3.png "Rendered by QuickLaTeX.com")
which are products of
![\[5,\text{ }are\text{ }105,\text{ }110,\text{ }\ldots \text{ }995.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-454f738a414f1eae5622c98d56b2c201_l3.png "Rendered by QuickLaTeX.com")
It plainly frames an arrangement in A.P.
Where, the initial term,
![\[a\text{ }=\text{ }105\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-31a960c468788f8ecd55fa6b89160af1_l3.png "Rendered by QuickLaTeX.com")
Normal contrast,
![\[d\text{ }=\text{ }5\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-743391f8823ec07d8f6815e8519cc72a_l3.png "Rendered by QuickLaTeX.com")
Presently,
![\[\begin{array}{*{35}{l}} a\text{ }+\text{ }\left( n\text{ }-1 \right)d\text{ }=\text{ }995 \\ 105\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\left( 5 \right)\text{ }=\text{ }995 \\ 105\text{ }+\text{ }5n\text{ }\text{ }5\text{ }=\text{ }995 \\ 5n\text{ }=\text{ }995\text{ }\text{ }105\text{ }+\text{ }5\text{ }=\text{ }895 \\ n\text{ }=\text{ }895/5 \\ n\text{ }=\text{ }179 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9d86f271e054eac7797a210f67c91b78_l3.png "Rendered by QuickLaTeX.com")
We know,
![\[{{S}_{n}}~=\text{ }n/2\text{ }\left[ 2a\text{ }+\text{ }\left( n-1 \right)d \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-417bcd262acac9d2f45730b4735d433e_l3.png "Rendered by QuickLaTeX.com")
In this manner, the amount of all regular numbers lying somewhere in the range of
which are products of
![\[5,\text{ }is\text{ }98450\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7720e329588c2adb25dd8e055baa465e_l3.png "Rendered by QuickLaTeX.com")
.