36] Find the sum of all natural numbers less than 100 which are divisible by 6
Answers
Answer:
Natural numbers less than 100 and divisible by 6:
6, 12, 18, 24, ... 96
Step-by-step explanation:
This forms an AP with 'a' = 6 and 'd' = 6, 'aₙ' = 96
⇒ aₙ = a + ( n - 1 ) d
⇒ 96 = 6 + ( n - 1 ) 6
⇒ 96 - 6 = ( n - 1 ) 6
⇒ 90 = ( n - 1 ) 6
⇒ 90/6 = ( n - 1 )
⇒ 15 = n - 1
⇒ n = 15 + 1 = 16 terms
Sum of 'n' terms of an AP:
⇒ Sₙ = n/2 [ a + aₙ ]
⇒ S₁₆ = 16/2 [ 6 + 96 ]
⇒ S₁₆ = 8 ( 102 )
⇒ S₁₆ = 816
Therefore sum of all natural numbers less than 100 which are divisible by 6 is 816.
Answer:
816
Step-by-step explanation:
This is a sum of an arithmetic sequence problem where the formula would be
S = (n/2) × (2a + (n−1)d) where n is the number of terms, a is the first number in the sequence and d would be the common difference between terms.
In your question, a, the first term, = 6.
n, the number of terms, = 16.
d, the common difference = 6.
So, S = (16/2) * (2*6 + (16–1)*6) = 8 * (12 + 15*6) = 8 * (12+90) = 8 * 102 = 816.