definite series!

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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