Question

find the sum of all natural numbers less than 100 which are divisible by 6​

Answer 1

6,12,18,...96

are all natural less than 100 and divisible by 6

first term =6

common difference= d = a2 -a1=12-6 =6

nth term = 9

6a+(n-1)d=96

6+(n-1)6=96

(n-1)6=96-6

(n-1)6=90n-1=90/6

n-1=15

n=15+1

n=16

sum = n/2[a+ nth term]

=16/2[6+96]

=8×102=816

Answer 2

 

Least multiple of 6 (natural number) below 100 = 6

Largest multiple of 6 (natural number) below 100 = 96

Let the multiples of 6 be in AP. It will be like 6, 12, 18, ......, 96.

⇒ 6 + 12 + 18 + ... + 96

⇒ 6(1 + 2 + 3 + ... + 16)

⇒ 6 [(16 × 17) / 2]                    {∵ 1 + 2 + 3 + ... + n = [n(n + 1)] / 2}

⇒ 6 × 8 × 17

⇒ 816

So the sum of natural numbers as multiples of 6 below 100 is 816!!!

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