Math, asked by madan7615, 11 months ago

Find the sum of the natural numbes between 1 and 101 which are divisible by 3

Answers

Answered by akshaybhushans
0

See natural numbers include all numbers in your career from1 to 101

Therefore we need to find multiples of 3 from 1 to 101 which are

3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99.

The sum is 1683

Hope it helps

Answered by parashuramnalla
0

Answer:

The integers between 1 and 101, which are divisible by 3, are 1683

Step-by-step explanation:

Find the sum of all natural number between 1 and 101 which are divisible by 3.

The integers from 1 to 101 are 2, 3, 4, ..........100

which are divisible by 3, are 3, 6, 9….........99.

This forms an A.P. with the first term 3 and common difference equal to 3.

99 = 3 + (n – 1) 3

99 = 3 + 3n - 3

99 =  3n + 0

n = 99 / 3

n = 33

Sn = ( n / 2 ) [ 2a + ( n – 1) d ]  

Sn = ( 33 / 2) [ 2 ( 3 )  + ( 33 – 1 ) 3 ]

Sn = ( 33 / 2) [ 6  + ( 32 ) 3 ]

Sn = ( 33 / 2) [ 6  + 96 ]

Sn = ( 33 / 2) [ 102 ]

Sn = ( 33  x 102 ) / 2

Sn = 33  x 51

Sn = 1683

The integers between 1 and 101, which are divisible by 3, are 1683

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