Question

Find the sum of all natural numbers lying between 100 and 1000

Answer 1

Answer:

A.P :- 101 , 102 , 103..........,....999

first term a = 101

common difference d = 102 - 101 = 1

last term = 999

then,

an = a + ( n -1) d

999 = 101 + ( n - 1) 1

999 - 101 = n - 1

898 = n - 1

898 + 1 = n

899 = n

then ,

$sn = \frac{n}{2} (2a + (n - 1)d) \\ \\ = \frac{899}{2} (2 \times 101 + (899 - 1)1) \\ \\ = \frac{899}{2} (202 +898) \\ \\ = \frac{899}{2} \times 1100 \\ \\ = 899 \times 550 \\ \\ = 494450$

Answer 2

Answer:

refers to the attachment