Question

find the sum of the first 12 positive integers divisible by 9​

Answer 1

The sum of the first 12 positive integers divisible by 9​ is 702.

The multiples of 9 are : 9,18,27,36 and so on.

We have to find the sum of the first 12 multiples.

We can see that the sequence is an Arithmetic Progression (A.P.), with common difference (d), between two terms equal to 9.

We know, sum of n terms in an AP is given as:

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

Here, n = 12, as we need to find the sum of first 12 multiples,

d = 9

a, the first term = 9.

Putting these values, we get:

S = (12/2)[18 + (11)(9)]

   = 6[18 + 99]

   = 6[117] = 702

This is the sum.

Answer 2

Answer:

702

Step-by-step explanation:

multiples of 9: 9,18,27,36 and so on

we know that sum of n term is an Ap is given as

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

n=12

d=9

first term is 9

putting the value, we get

s=12/2(18+(11)(9))

6(18+99)

6(117)

702