Question

Find the sum of first five positive consecutive integers divisible by 3​

Answer 1

Step-by-step explanation:

Given:-

First five positive consecutive integers divisible by 3

To find:-

Find the sum of first five positive consecutive integers divisible by 3 ?

Solution:-

The numbers are divisible by 3

= 3,6,9,...

The general form of the numbers which are divisible by 3

= 3n

The first five positive consecutive integers divisible by 3

= 3,6,9,12,15

Their sum = 3+6+9+12+15 =45

or

The first five. positive integers which are divisible by 3

= 3,6,9,12,15

We know that

The sum of first n terms in an AP

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

=>S5 = (5/2)[2(3)+(5-1)3]

=> S5 = (5/2)[6+4(3)]

=> S5 = (5/2)(6+12)

=> S5 = (5/2)(18)

=> S5 = 5×9

=> S 5 = 45

Answer:-

The sum of first five positive consecutive integers divisible by 3 = 45

Used formulae:-

• The sum of first n terms in an AP
• Sn = (n/2)[2a+(n-1)d]

• The general form of the numbers which are divisible by 3 = 3n