Question

The sum of first n natural numbers is given by 1/2 n^2 + 1/2 n. 1) Find the sum of first 11 natural numbers. 2) find the sum of natural numbers from 11 to 30. NO WRONG ANSWERS!!!

Answer 1

Answer:

$\sf{(1) \ The \ sum \ of \ first \ 11 \ natural}$

$\sf{natural \ is \ 66.}$

$\sf{(2) \ The \ sum \ of \ natural \ numbers \ from}$

$\sf{11 \ to \ 30 \ is \ 410.}$

Given:

$\sf{The \ sum \ of \ first \ n \ natural}$

$\sf{numbers \ is \ given \ by}$

$\sf{\dfrac{1}{2}\times \ n^{2}+\dfrac{1}{2}\times \ n}$

To find:

$\sf{1) \ The \ sum \ of \ 11 \ natural \ numbers.}$

$\sf{2) \ The \ sum \ of \ natural \ numbers \ from}$

$\sf{11 \ to \ 30.}$

Solution:

$\sf{(1)}$

$\sf{S_{n}=\dfrac{1}{2}\times \ n^{2}+\dfrac{1}{2}\times \ n}$

$\sf{Put \ n=11 \ we \ get,}$

$\sf{\therefore{S_{11}=\dfrac{1}{2}\times11^{2}+\dfrac{1}{2}\times11}}$

$\sf{\therefore{S_{11}=\dfrac{121}{2}+\dfrac{11}{2}}}$

$\sf{\therefore{S_{11}=\dfrac{121+11}{2}}}$

$\sf{\therefore{S_{11}=\dfrac{132}{2}}}$

$\sf{\therefore{S_{11}=66}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ first \ 11 \ natural}}}$

$\sf\purple{\tt{natural \ is \ 66.}}$

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$\sf{(2)}$

$\sf{Sum \ of \ natural \ numbers \ from \ 11 \ to \ 30}$

$\sf{can \ given \ by \ S_{30}-S_{10}}$

$\sf{=(\dfrac{1}{2}\times30^{2}+\dfrac{1}{2}\times30)-(\dfrac{1}{2}\times10^{2}+\dfrac{1}{2}\times10)}$

$\sf{=\dfrac{1}{2}(30^{2}+30)-\dfrac{1}{2}(10^{2}+10)}$

$\sf{=\dfrac{1}{2}(900+30)-\dfrac{1}{2}(100+10)}$

$\sf{=\dfrac{1}{2}(930-110)}$

$\sf{=\dfrac{1}{2}(820)}$

$\sf{=410}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ natural \ numbers \ from}}}$

$\sf\purple{\tt{11 \ to \ 30 \ is \ 410.}}$