Question 2: Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Answer:
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.
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