three numbers of a geometric progression whose sum is 28 and whose product is 512. find the first three numbers

the problem above, i have attempted it and arrived at 8,16,32

am i correct? thanks

Printable View

- May 21st 2012, 03:06 AMlawochekeldefinite series!
three numbers of a geometric progression whose sum is 28 and whose product is 512. find the first three numbers

the problem above, i have attempted it and arrived at 8,16,32

am i correct? thanks - May 21st 2012, 03:27 AMPlatoRe: definite series!
- May 21st 2012, 03:53 AMProve ItRe: definite series!
As Plato has shown, your solution can not possibly be right.

You have

$\displaystyle \displaystyle \begin{align*} t_n + r\,t_n + r^2t_n &= 28 \\ (r^2 + r + 1)t_n &= 28 \end{align*}$

and

$\displaystyle \displaystyle \begin{align*} t_n \cdot r\,t_n \cdot r^2t_n &= 512 \\ r^3t_n &= 513 \end{align*}$

Try to solve these two equations simultaneously to find $\displaystyle \displaystyle \begin{align*} r \end{align*}$ and $\displaystyle \displaystyle \begin{align*} t_n \end{align*}$. From there you can find the following two terms. - May 21st 2012, 12:55 PMbjhopperRe: definite series!
Try this

3 terms a,ar.ar^2 Product = 512 = a^3 r^3 a = 8/r second term =8 third term = 8r

add the 3 terms whose sum =28 . This will produce a quadratic equation giving the r value