Question

find the sum of the first 30 positive integer divisible by 6
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Answer 1

Step-by-step explanation:

6,12,18,24,30

a=6

d=6

n=30

Sn=?

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

30/2(2×6+(30-1)6

15(12+(29)6

15(12+174)

15(186)

2790

Answer 2

Step-by-step explanation:

Given:-

First 30 positive integers divisible by 6

To find:-

Find the sum of the first 30 positive integers which are divisible by 6 ?

Solution:-

The list of positive integers

= 1,2,3,4,...

The list of positive integers which are divisible by 6

= 6,12,18,24,30,...

The list of first 30 positive integers which are divisible by 6

= 6,12,18,...(30 terms)

First term = a = 6

Common difference = 12-6=6

Since the common difference is same throughout the series

6,12,18,...are in the AP

Now we have to find the sum of first 30 positive integers which are divisible by 6

= 6+12+18+...+(30 terms)

We know that

The sum of first n terms in an AP

= Sn = (n/2)[2a+(n-1)d]

We have

a = 6

d = 6

n = 30

Now,

S 30 = (30/2)[2(6)+(30-1)(6)]

=> S 30 = (15)[12+29(6)]

=> S 30 = (15)(12+174)

=> S 30 = 15(186)

=> S 30 = 2790

Answer:-

The sum of first 30 positive integers which are divisible by 6 is 2790

Used formulae:-

The sum of first n terms in an AP

Sn = (n/2)[2a+(n-1)d]

• n = number of terms
• a = First term
• d = Common difference