Question

Find sum of the first 40 positive integer divisible by 6​

Answer 1

first look at the number which is divisible by 6

it is itself 6 now look at the next number, which is 12 now look the next number which is 18.

now it becomes an ap having a common difference of 6

6,12,18.......

a=6,d=6,n=40

sn=n/2(2a+n-1(d))

sn=20(12+39(6)

thankyou

Answer 2

Answer:

Step-by-step explanation:

The first 40 positive integers divisible by 6 are 6, 12, 18, 24, 30, ......., 240.

Now, we have to find the value of the sum of the first 40 integers divisible by 6.

Hence, we have to get

6+ 12+ 18+ 24+ 30+ ......... +240

Taking 6 as common from all the terms, we get,

6(1+ 2+ 3+ 4+ ......+ 40)

Now, the sum within the brackets is the sum of the first 40 natural numbers.

Hence, 6+ 12+ 18+ 24 +30 + ....... +240

= 6(1+ 2+ 3+ 4+ ......+ 40)

= 6 [40(40+1)/2] {Sum of first n natural numbers is given by n(n+1)/2}

=4920 (Answer)