# definite series!

• May 21st 2012, 03:06 AM
lawochekel
definite 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 AM
Plato
Re: definite series!
Quote:

Originally Posted by lawochekel
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

Well $8+16+32\ne 28$. What do you think?
• May 21st 2012, 03:53 AM
Prove It
Re: definite series!
Quote:

Originally Posted by lawochekel
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

As Plato has shown, your solution can not possibly be right.

You have

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

and

\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 \begin{align*} r \end{align*} and \displaystyle \begin{align*} t_n \end{align*}. From there you can find the following two terms.
• May 21st 2012, 12:55 PM
bjhopper
Re: 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