Question

Find the sum of all 3 digit natural multiples of 6.

Answer 1

Dear student,

Answer: Sum of all 3 digit natural multiples of 6 is 82,350.

Solution:

Three digits natural multiples of 6 would be 102 , 108, 114, 120 ... 996

From the given numbers we can write that it is an AP.

a = 102

d = 6

to calculate the sum of all terms we had to calculate the total number of element in the AP, from its last term

let the last term be nth term

996 = 102 +(n-1) 6

996-102 = (n-1) 6

(n-1)6 = 894

n-1 = 894/6

n-1 = 149

n= 149+1

n = 150

Sum of all 3 digit natural multiples of 6 are = n/2( a+l)

= 150/2( 102+996)

= 150/2(1098)

= 150(549)

= 82,350

sum of all 3 digit natural multiples of 6 is 82,350.

Hope it helps you.

Answer 2

First 3-digit number divisible by 6 = 102.

Last 3-digit number divisible by 6 = 996.

= > 102 + 108 + 114... +996.

Clearly, this series is in AP.

First term a = 102, Common difference d = 108 - 102 = 6, Last term tn = 996.

Now,

We know that tn = a + (n - 1) * d

= > 996 = 102 + (n - 1) * 6

= > 996 = 102 + 6n - 6

= > 996 = 96 - 6n

= > 996 + 96 = 6n

= > 1092 = 6n

= > n = 150.

Now,

Sum of the required series sn = n/2[a + l]

= > (150/2)[102 + 996]

= > 75[1098]

= > 82350.

Hope this helps!