Question

find the sum of all natural numbers between hundred and thousand which are multiple of 5​

Answer 1

Answer:

as we know a(n-1)d

so ,100(900-1)5

=449500

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

First term (a) = 105

Common difference

d = 5

Using Formula :-

Here,

a + (n - 1)d = 995

105 + (n - 1)5 = 995

(n - 1)5 = 995 - 105

(n - 1)5 = 890

$\tt{\rightarrow(n-1)=\dfrac{890}{5}}$

n - 1 = 178

n = 178 + 1

n = 179

$\tt{\rightarrow S_{n}=\dfrac{n}{2}[2a+(n-1)d]}$

$\tt{\rightarrow S_{179}=\dfrac{179}{2}[2(105)+(179-1)5]}$

$\tt{\rightarrow S_{179}=\dfrac{179}{2}[2(105)+(178)5]}$

= 179[105+(89)5]

= 179[105+445]

= 179 × 550

= 98450

$\Large{\boxed{\sf\:{Hence\;the\;sum\;is\;98450}}}$