Question

Find the sum of all natural numbers upto 100 which are divisible by 3.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\sf Sum \; of \; all \; natural \; Numbers\; upto\; 100\; which\; are\; Divisible\; by \;3 =1683$

Step-by-step explanation:

The Natural Numbers upto 100 which are divisible by 3 are :- 3,6,9,12....99 .

So , AP = 3,6,9,12.......99

$\bigstar$ Here in this AP :- The comman difference (d) = 3 , the first term a = 3 and the last term or a(n) = 99 .

So,

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

99 = 3 + (n-1)3

99 = 3 + 3n - 3

99 = 3n

n = 99/3 = 33

So the AP has a total of 33 Terms in it .

Sum of an AP , S(n) = n/2( 2a+(n-1)d)

S(33) = $\dfrac{33}2$ ( 2×3+ (33-1)3)

S(33) = $\dfrac{33}2$ (6+32×3)

S(33) = $\dfrac{33}2$ (6+96)

S(33) = $\dfrac{33}2$(102)

S(33) = 33× 51 = $\underline{\underline{{1683}}}$