1. ## 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

2. ## Re: definite series!

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 $\displaystyle 8+16+32\ne 28$. What do you think?

3. ## Re: definite series!

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 \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.

4. ## 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