Question

Find the sum of the first 1000 positive integers.

Answer 1

In this problem, the positive integers are in A.P., hence, we need to apply the arithmetic progression formula to obtain the sum of the first 1000 positive integers.

Complete step-by-step answer:
The positive integers start from 1.
The series of the positive integers starting from 1, and end at 1000 is shown below.
$1,2,3,4,5, \ldots \ldots ,1000$
The total number of terms $n$ in the series are 1000.
First number $a$ of the series is 1 and the common difference $d$ is 1.
The formula for the sum $S$ of $n$ terms in A.P. is shown below.
S = \dfrac{n}{2}\left\\{ {2a + \left( {n - 1} \right)d} \right\\}
Substitute 1000 for$n$, 1 for $a$ and for $d$ in the above equation.

\,\,\,\,S = \dfrac{{1000}}{2}\left\\{ {2\left( 1 \right) + \left( {1000 - 1} \right)1} \right\\} \\\ \Rightarrow S = 500\left\\{ {2 + 999} \right\\} \\\ \Rightarrow S = 500\left\\{ {1001} \right\\} \\\ \Rightarrow S = 500500 \\\

Thus, the sum of the first 1000 positive integers is 500500.

Note: The given positive integers are in arithmetic progression. Apply the formula for the sum of $n$ terms in arithmetic progression.