Question

Find the sum of all numbers between 100 and 500 which are divisible by 3.

Answer 1

Between 100 & 500 numbers which are Divisible by 3 = 102,105,..........,498
First term. = 102
Common difference d= 3
n th term = a+(n -1)d
=102+(n-1) x3=498
(n-1)3=498-102=396
n-1=396/3=132
n=132+1=133
Sum of all numbers = Sn =n/2[a+l]

=S133=133/2[102+498]=133/2[600]=
133x300=39900

Answer 2

Answer:

$\boxed{\sf The \; sum \;of \;all\; numbers \;between\; 100 \;and \;500 \;which\; are \;divisible\; by \;3\; is \: 39900}$

Step-by-step explanation:

The numbers between 100 and 500 , that are divisible by three are :-

102,105,108,111, ....... 498.

When we observe this sequence we can see that this forms an AP , whose

First term (a) = 102

Common Difference (d) = a2-a1 = 105-102 = 3

Last Term (an) = 498

∵ a = 102 , d = 3 and a(n) = 498

a(n) = a +(n-1)d

498 = 102 + ( n-1)3

498 = 102 + 3n -3

498 - 102 = 3n-3

396+3 = 3n

399 = 3n

$n = \dfrac{399}{3} = \underline{\underline{{133}}}$

The AP has 133 terms .

Sum

$S(n) = \dfrac{n}{2} (a+a(n))$

$S(n) = \dfrac{133}{2} (102+498)$

$S(n) = \dfrac{133}{2} * 600$

$S(n) = 133 * \dfrac{600}{2}$

$S(n) = 133 \; *\;300$

$S(n) = \underline{\underline{{39900}}}$