Question

find the sum of the first 40 positive integers divisible by 6

Answer 1

first 40 positive integers divisible by six are
6,12,18,24,30,36
so,there sum is equal to 6+12+18+24+30+36=126
it's ur answer

Answer 2

Answer:

4920

Step-by-step explanation:

The positive integers divisible by 6 are :

6, 12 , 18, ....

Since the common difference between each number is same, the given number is in an AP.

We need to find the sum of first 40 integers,

Common difference (d) = 6

Number of terms (n) = 40

First term (a) = 6

Here, we can use formula ;

$\bf S_n = \frac{n}{2} [2a + (n - 1)d]$

$\bf S_{40} = \frac{40}{2} [2 \times 6 + (40 - 1)6] \\ \\ \bf S_{40} = 20 [12 + 234] \\ \\ \bf S_{40} = 20 \times 246 \\ \\ \red{\boxed{\bf S_{40} = 4920}}$

Hence, the sum of 40 integers divisible by 6 is 4920.