Question

Find the sum of all natural numbers between 400 and 500 which are divisible by 3

Answer 1

Here’s the answer to the question

Answer 2

Answer:

Step-by-step explanation:

Here first term is a = 402

Last term l = 459

common difference d = 3

Number of terms n :

l = a + ( n- 1 ) d

459 = 402 + ( n - 1 ) 3

=> n= 57/3 + 1

n = 19 + 1

=>n = 20

Sum of terms = n/2 (a + l )

= 20/2 ( 402 + 459 )

= 10 × 861

= 8610

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