Question

What is the sum of natural numbers from 11 to 40?
(1) 820 (2) 765 (3) 355 (4) 610​

Answer 1

$\textsf{Concept:}$

$\textsf{The sum of first n natural numbers}$

$\mathsf{1+2+3.....+n=\displaystyle\frac{n(n+1)}{2}}$

$\mathsf{Now, 11+12+13+...........+40}$

$\mathsf{=(1+2+3........+10+11+....+40)-(1+2+....+10)}$

$\mathsf{=\displaystyle\frac{40{\times}41}{2}-\frac{10{\times}11}{2}}$

$\mathsf{=20{\times}41-5{\times}11}$

$\mathsf{=820-55}$

$\mathsf{=765}$

$\therefore\boxed{\textsf{11+12+13....+40=765}}$

$\implies\,\textsf{option (2) is correct}$

Answer 2

The sum of natural numbers from 11 to 40 is 765.

Step-by-step explanation:

We have to find the sum of natural numbers from 11 to 40.

This can be considered as an AP as the common difference is 1 for this series.

As we know that, the sum of n terms formula of an AP is given by;

                      $S_n=\frac{n}{2} [2a+(n-1)d]$

where, a = first term of an AP = 11

           d = common difference = 1

            n = number of terms = 40 - 11 + 1 = 30

Now,   $S_3_0 = \frac{30}{2} [(2\times 11)+(30-1)\times 1]$

           $S_3_0 = 15 \times [22+29]$

           $S_3_0 = 15 \times 51$

           $S_3_0 =765$

Hence, the sum of natural numbers from 11 to 40 is 765.