Question

Find the sum of all the netural numbers between 100 and 1000

Answer 1

$\sf\large\underline{Solution:}$

• At first calculate all the natural number that comes between 100 and 1000 than calculate their sum .This can be solved by AP easily]

$\rm{\implies \therefore\:AP are 101,\:102,\:103.........999}$

$\rm{\implies First\: term(a)=101}$

$\rm{\implies Common\: difference=102-101=1}$

$\rm{\implies Last\:term\:T_{n}=999}$

$\sf\large\underline{Formula\: used:}$

$\rm{\implies T_{n}=a+(n-1)d}$

$\rm{\implies 999=101+(n-1)1}$

$\rm{\implies 999-101=(n-1)}$

$\rm{\implies n-1=898}$

$\rm{\implies n=900}$

$\rm{\implies \therefore\: Number\:n\:terms=900}$

• Now calculate the sum here]

$\rm{\implies S_{n}=\dfrac{n}{2}\bigg(2a+(n-1)d\bigg)}$

$\rm{\implies S_{900}=\dfrac{900}{2}\bigg(2\times\:101+(101-1)1\bigg)}$

$\rm{\implies S_{900}=\dfrac{900}{2}\bigg(202+100)\bigg)}$

$\rm{\implies S_{900}=\dfrac{900}{2}\times\:302}$

$\rm{\implies S_{900}=450\times\:302}$

$\rm{\implies S_{900}=135900}$

$\sf\large{Hence,}$

$\rm{\implies The\:sum\:of\:all\: natural\:namber=135900}$

Answer 2

105,110......................995

a = 105
d = 5
l = 995
5n = ?, n = ?.

$\bf \: l = a + (n - 1)d$
$\bf1000 = 100 + (n - 1)5$

$\bf \cancel{900} = \cancel{5}(n - 1)5$

$\bf \: n - 1 = 180$

$\bf \boxed{\red{n = 181}}$

$\bf \: Sn = \frac{n}{2} [2 \times 9 + (n - 1)d]$

$\: \: \: \: \: \: \: \: \: \: \to \bf \: \frac{181}{2} [2 \times 100 + (180) \times 5]$

$\: \: \: \: \: \: \: \: \: \: \: \to \bf181[100 + 90 \times 5]$

$\: \: \: \: \: \: \: \: \: \: \to \bf181 \times 550$

$\boxed{ \red{ \bf5n \: = \: 99550}}$