Question

The sum of three consecutive terms of an arithmetic progression is 18 and there products is 120 find the numbers

Answer 1

Answer:

Step-by-sterkenekndiendsxp explanation:

Answer 2

Answer:

10, 6 & 2

Step-by-step explanation:

let the numbers be

x, x + d, x + 2d

here, d is the difference between the progression of AP

now,

given that sum of the three consecutive terms of AP is 18

& their product is 120

putting these values on we get

x + (x+d) + (x+2d) = 18

or

3x + 3d = 18

or

3(x+d) = 18

or

x+d = 6......(i)

or d = 6-x.....(ii)

now,

x × (x+d) × (x+2d) = 120

now putting value of x+d from equation (i) we get

x × 6 × (x+2d) = 120

or

6x(x+2d) = 120

or

6x^2 + 12xd = 120

or

6(x^2 + xd) = 120

or

x^2 + xd = 60

now putting the value of d from equation (ii) we get

x^2 + x(6-x) = 60

or

x^2 + 6x- x^2 = 60

or

6x = 60 { +x^2 is cancelled with -x^2 }

or x = 10

so

d = 6-10 = (-4)

first number = 10

second number = 10 + (-4) = 10-4 = 6

third number = 10 + (2×-4) = 10-8 = 2