MATH SOLVE

7 months ago

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

[tex]a_n=a(r)^{n-1}[/tex]

so the first 3 terms are

[tex]a_1=a[/tex]

[tex]a_2=a(r)[/tex]

[tex]a_2=a(r)^2[/tex]

the sum is -7/10

[tex]\frac{-7}{10}=a+ar+ar^2[/tex]

and their product is -1/125

[tex]\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3[/tex]

from the 2nd equation we can take the cube root of both sides to get

[tex]\frac{-1}{5}=ar[/tex]

note that a=ar/r and ar²=(ar)r

so now rewrite 1st equation as

[tex]\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r[/tex]

subsituting -1/5 for ar

[tex]\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r[/tex]

which simplifies to

[tex]\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}[/tex]

multiply both sides by 10r

-7r=-2-2r-2r²

add (2r²+2r+2) to both sides

2r²-5r+2=0

solve using quadratic formula

for [tex]ax^2+bx+c=0[/tex]

[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

so

for 2r²-5r+2=0

a=2

b=-5

c=2

[tex]r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}[/tex]

[tex]r=\frac{5 \pm \sqrt{25-16}}{4}[/tex]

[tex]r=\frac{5 \pm \sqrt{9}}{4}[/tex]

[tex]r=\frac{5 \pm 3}{4}[/tex]

so

[tex]r=\frac{5+3}{4}=\frac{8}{4}=2[/tex] or [tex]r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}[/tex]

use them to solve for the value of a

[tex]\frac{-1}{5}=ar[/tex]

[tex]\frac{-1}{5r}=a[/tex]

try for r=2 and 1/2

[tex] a=\frac{-1}{10}[/tex] or [tex]a=\frac{-2}{5}[/tex]

test each

for a=-1/10 and r=2

a+ar+ar²=[tex]\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}[/tex]

it works

for a=-2/5 and r=1/2

a+ar+ar²=[tex]\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}[/tex]

it works

both have the same terms but one is simplified

the 3 numbers are [tex]\frac{-2}{5}[/tex], [tex]\frac{-1}{5}[/tex], and [tex]\frac{-1}{10}[/tex]

geometric sequences can be represented as

[tex]a_n=a(r)^{n-1}[/tex]

so the first 3 terms are

[tex]a_1=a[/tex]

[tex]a_2=a(r)[/tex]

[tex]a_2=a(r)^2[/tex]

the sum is -7/10

[tex]\frac{-7}{10}=a+ar+ar^2[/tex]

and their product is -1/125

[tex]\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3[/tex]

from the 2nd equation we can take the cube root of both sides to get

[tex]\frac{-1}{5}=ar[/tex]

note that a=ar/r and ar²=(ar)r

so now rewrite 1st equation as

[tex]\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r[/tex]

subsituting -1/5 for ar

[tex]\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r[/tex]

which simplifies to

[tex]\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}[/tex]

multiply both sides by 10r

-7r=-2-2r-2r²

add (2r²+2r+2) to both sides

2r²-5r+2=0

solve using quadratic formula

for [tex]ax^2+bx+c=0[/tex]

[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

so

for 2r²-5r+2=0

a=2

b=-5

c=2

[tex]r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}[/tex]

[tex]r=\frac{5 \pm \sqrt{25-16}}{4}[/tex]

[tex]r=\frac{5 \pm \sqrt{9}}{4}[/tex]

[tex]r=\frac{5 \pm 3}{4}[/tex]

so

[tex]r=\frac{5+3}{4}=\frac{8}{4}=2[/tex] or [tex]r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}[/tex]

use them to solve for the value of a

[tex]\frac{-1}{5}=ar[/tex]

[tex]\frac{-1}{5r}=a[/tex]

try for r=2 and 1/2

[tex] a=\frac{-1}{10}[/tex] or [tex]a=\frac{-2}{5}[/tex]

test each

for a=-1/10 and r=2

a+ar+ar²=[tex]\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}[/tex]

it works

for a=-2/5 and r=1/2

a+ar+ar²=[tex]\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}[/tex]

it works

both have the same terms but one is simplified

the 3 numbers are [tex]\frac{-2}{5}[/tex], [tex]\frac{-1}{5}[/tex], and [tex]\frac{-1}{10}[/tex]