 Question 2

# Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

This sequence forms an A.P.

Here, first term, a = 105

Common difference, d = 5

Here,

\begin{align} a + (n - 1)d = 995 \end{align}

\begin{align} => 105 + (n - 1)5 = 995 \end{align}

\begin{align} =>  (n - 1)5 = 995 - 105 = 890 \end{align}

\begin{align} => n -1 = 178 \end{align}

\begin{align} => n = 179 \end{align}

\begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}

\begin{align} \therefore S_n = \frac {179}{2}\left[2 × (105) + (179 -1)×(5)\right]\end{align}

\begin{align} = \frac {179}{2}\left[2(105) + (178)(5)\right]\end{align}

\begin{align} = 179\left[105 + (89)5\right]\end{align}

\begin{align} = (179)\left[105 + 445\right]\end{align}

\begin{align} =179 × 550 \end{align}

\begin{align} = 98450 \end{align}

Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.