Question

Find the sum of the first 1000 positive integers.

Answer 1

Hint: 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.