## find the sum of the first 30 positive integer divisible by 6​

Question

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

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5 months 2021-07-08T01:29:24+00:00 2 Answers 0 views 0

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

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