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Class 11
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find the sum of all natural numbers lying between...

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

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