Question

5. The numbers x, 17/12 and y are the first three terms of an arithmetic progression.

The numbers x, 2/3 and y are three consecutive terms of a geometric progression. Determine the values of x and y.

Answer 1

`(1)/(6)$ and $(8)/(3)$ or $ (8)/(3)$ and $ (1)/(6)$ respectively.`

Explanation:

The numbers x,
`(17)/(12)` and y are the first three terms of an arithmetic progression.

For three consecutive terms, a, b and c of an arithmetic sequence:

The arithmetic mean,
`b=(a+c)/(2)`

Therefore:

`(17)/(12)=(x+y)/(2)\\12(x+y)=17*2\\12(x+y)=34\\x+y=(34)/(12)\\\\x= (34)/(12)-y`

The numbers x,
`\frac23` and y are three consecutive terms of a geometric progression.

For three consecutive terms, a, b and c of a geometric sequence:

The geometric mean,
`b=√(ac)`

Therefore:

`\frac23=√(xy)\\xy=(\frac23)^2\\xy=\frac49`

Substituting
`x= (34)/(12)-y` derived above, we have:

`((34)/(12)-y)y=\frac49\\\\(34)/(12)y-y^2=\frac49\\\\y^2-(34)/(12)y+\frac49=0`

This is a quadratic equation which we can then solve for y.

Using a calculator, we obtain:

`y=(1)/(6)$ or $y= (8)/(3)`

`When\: y=(1)/(6), x= (34)/(12)-(1)/(6)=(8)/(3)\\\\When\: y=(8)/(3), x= (34)/(12)-(8)/(3)=(1)/(6)`

Therefore, the values of x and y are:

`(1)/(6)$ and $(8)/(3)$ or $ (8)/(3)$ and $ (1)/(6)$ respectively.`