Question

The sum of three consecutive multiples of 4 is 36. find the multiples​

Answer 1

Answer:

Let the three consecutive multiples of 11 are 11x, 11x+11 and 11x+22

Their sum is 363

11x+11x+11+11x+22=363

33x+33=363

33x=363−33

33x=330

x=

33

330

x=10

First multiple =11x=11×10=110

Second multiple =11x+11=110+11=121

Third multiple =11x+22=110+22=132

Answer 2

Answer:

6,12,18

Step-by-step explanation:

let 3 consq. mult. be 4x,8x,12x

4x+8x+12x=36 ( given)

24x=36

x=36/24

x=3/2

Substituting in x

4x=4(3/2)=6

8x=8(3/2)=12

12x=12(3/2)=18

Proving

18+12+6=36

36=36

LHS=RHS