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