Question

The product of three numbers in gp is1000. If 6 is added to the second number and 7 is added to the third number, we get an AP. Find the numbers.​

Answer 1

Answer:

Let the numbers be a/r,a,ar.

From question

a/r × a ×ar =1000

a^3 =1000

a = 10.

Then

a/r ,a + 6,ar+7 are in ap

So common difference is equal

(a+6)-(a/r)=(ar+7)-(a+6)

Solving equation by substituting the value of a as 10 giver the equation

2r^2-5r+2=0

So by solving this equation we get r is 1/2 or 2.

By substituting values of a ,r in a/r,a,ar gives the numbers

20,10,5

Answer 2

Answer:

5,10,20

Step-by-step explanation:

Let the three numbers be a/r , a , ar

Then, a/r×a×ar=1000

a^3=1000

a=10

Given, 10/r , 16 , 10r+7 are in A.P

32=10/r+10r+7

By solving we get,

2r^2+5r+2=0

r=2,1/2

so the numbers are 5 , 10 , 15.