Find the sum of all natural number lying between 100 and 1000 which ar

1.	Find the sum of all natural number lying between 100 and 1000 which are multiples of 5.
Answer» $\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$ Natural numbers are FOLLOWS 105,110,115,120,...............,995 Here we have First term(a) = 105 Common difference (d) = 110 - 105 = 5 Using formula we have ${\boxed{\sf\:{a_{n}=a+(n-1)d}}}$ Hence, a + (n - 1)d = 995 Putting the VALUES :- 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 Now, ${\boxed{\sf\:{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}$ $\tt{\rightarrow S_{179}=\dfrac{179}{2}[2(105)+(179-1)(5)]}$ $\tt{\rightarrow S_{179}=\dfrac{179}{2}[2(105)+(178)(5)]}$ = 179[105 + (89) × 5] = 179[105 + 445] = 179 × 550 = 98450

Find the sum of all natural number lying between 100 and 1000 which are multiples of 5.

Answer»

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

Natural numbers are FOLLOWS

105,110,115,120,...............,995

Here we have

First term(a) = 105

Common difference (d) = 110 - 105 = 5

Using formula we have

${\boxed{\sf\:{a_{n}=a+(n-1)d}}}$

Hence,

a + (n - 1)d = 995

Putting the VALUES :-

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

Now,

${\boxed{\sf\:{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}$

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

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

= 179[105 + (89) × 5]

= 179[105 + 445]

= 179 × 550

= 98450