Q:

# The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to represent the first three terms, respectively.] The three number are __, __, and __

Accepted Solution

A:
ooh, fun

geometric sequences can be represented as
$$a_n=a(r)^{n-1}$$
so the first 3 terms are
$$a_1=a$$
$$a_2=a(r)$$
$$a_2=a(r)^2$$

the sum is -7/10
$$\frac{-7}{10}=a+ar+ar^2$$
and their product is -1/125
$$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$$

from the 2nd equation we can take the cube root of both sides to get
$$\frac{-1}{5}=ar$$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$$
subsituting -1/5 for ar
$$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$$
which simplifies to
$$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$$
multiply both sides by 10r
-7r=-2-2r-2r²
2r²-5r+2=0
for $$ax^2+bx+c=0$$
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$$
$$r=\frac{5 \pm \sqrt{25-16}}{4}$$
$$r=\frac{5 \pm \sqrt{9}}{4}$$
$$r=\frac{5 \pm 3}{4}$$
so
$$r=\frac{5+3}{4}=\frac{8}{4}=2$$ or $$r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$$

use them to solve for the value of a
$$\frac{-1}{5}=ar$$
$$\frac{-1}{5r}=a$$
try for r=2 and 1/2
$$a=\frac{-1}{10}$$ or $$a=\frac{-2}{5}$$

test each
for a=-1/10 and r=2
a+ar+ar²=$$\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$$
it works

for a=-2/5 and r=1/2
a+ar+ar²=$$\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$$
it works

both have the same terms but one is simplified

the 3 numbers are $$\frac{-2}{5}$$, $$\frac{-1}{5}$$, and $$\frac{-1}{10}$$