Question

Find the sum of all the integers from 1 to 1000
$
  (a){\text{ 5050}} \\
  (b){\text{ 50500}} \\
  (c){\text{ 500500}} \\
  (d){\text{ 50050}} \\
$

Answer 1

Hint: In this question form a series all the integers from 1, 2, 3……………..1000. It forms an arithmetic progression. Use the direct formula of the sum of n terms in A.P that is ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$. The number of terms in this series will be 1000 only.

Complete step-by-step solution:
As we know (1, 2, 3, 4………………………, 1000) forms an A.P with first term (a = 1), common difference (d = (2 -1) = (3 - 2) =1) and the number of terms (n =1000).
Now we know the formula of sum of an A.P which is given as
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ Where (a is first term, d is common difference and n is number of terms).
So the sum of all integers’ numbers from 1 to 1000 is
$ \Rightarrow {S_{1000}} = \dfrac{{1000}}{2}\left( {2\left( 1 \right) + \left( {1000 - 1} \right)1} \right)$
Now simplify the above equation we have,
$ \Rightarrow {S_{1000}} = 500\left( {2 + 999} \right) = 500\left( {1001} \right) = 500500$
So this is the required sum of all natural numbers from 1 to 1000.
So this is the required answer.
Hence option (C) is correct.

Note: Terms are said to be A.P if the common difference that is the difference between the consecutive terms remains constant throughout the series. An integer is a number that can be written without a fractional component. It is advised to remember the general formula for terms in series like A.P and G.P. The total numbers can also be evaluated using the concept that any nth term in A.P can be written as ${a_n} = a + (n - 1)d$, so if we substitute 1000 in place of ${a_n}$ and the values of a and d, the total number of terms up to 1000 can be calculated and it will eventually 1000 only.