Thread: 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

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 . What do you think?

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



and



Try to solve these two equations simultaneously to find and . From there you can find the following two terms.

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