Question

The product of three consecutive terms in a geometric sequence is -1000 , and their sum is 15. Find the common ratio. (There are two answers. Separate them by commas)

Suggestion: Denote the terms by a/r, a, and ar.

Answer 1

is this for calculus II?

Answer 2

this a pretty cool problem, but the hint gives it away

Answer 3

product is
\[\frac{a}{r}\times a\times ar=a^3=-1000\] so you know
\[a=-10\]

Answer 4

then to solve for r you put
\[\frac{-10}{r}-10-10r=15\] so
\[-\frac{10}{r}-10r=25\] then turn into a quadratic and solve for r

Answer 5

yeah stupid hint...

Answer 6

maybe a stupid hint but it sure helped me because i wrote
\[a\times ar \times ar^2=-1000\] and got stuck right away, before looking at the hint. reading is a good skill

Answer 7

I see! So would it be -10, 25?
I got -10,5,20 but that isn't it.

Answer 8

no all we got was
\[a=-10\]

Answer 9

you still have to find "r"

Answer 10

see my post above as to how to find it

Answer 11

geometric sequence look like \[(\frac{a}{n})^{z}\] right?

Answer 12

i got
\[r=-\frac{1}{2}\text { or } r=-2\]

Answer 13

maybe \[\frac{n}{a}\]

Answer 14

I got r=-2 as well.

Answer 15

did you solve
\[-\frac{10}{r}-10-10r=15\]?

Answer 16

yep!