# The sum of three numbers in G.P. is 13/12 and their product is -1. Which are these numbers ?

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Let a1,a2,a3 be part of a G.P

==> a1 + a2 + a3 = 13/12.... (1)

==> a1*a2*a3 = -1............(2)

But:

a1 = a1

a2= a1*r

a3 = a1*r^2

==> a1+ a2 + a3 = a1 + a1*r + a1*r^2

==> a1 + a1r + a1r^2 = 13/12........(1)

Also:

a1*a2*a3 = -1

==> a1*a1*r *a1^r^2 = a1^3 * r^3 = (a1*r)^3

==> (a1*r)^3 = -1

==> a1*r = -1

==> a1= -1/r........(2)

Now substitute in (1):

==> a1 + a1*r + a1*r^2 = 13/12

==? -1/r  -1 + -r = 13/12

==> (-1 -r -r^2)/r = 13/12

==> Cross multiply:

==> 13r = 12(-r^2 -r -1)

==> 13r = -12r^2 - 12r -12

==> 12r^2 + 25r + 12 = 0

==> r 1= -25 + sqrt(625 - 4*144) /24

= [-25 + sqrt(49)]/24

= -25 + 7/24= -18/24 = -3/4

==> r1= -3/4

==> r2= -25 - 7/24 = -32/24 = -4/3

Then we have two solutions:

When r= -3/4

==> a1= 4/3

a2= 4/3*(-3/4) = -1

a3= 4/3*(-3/4)^2 = 3/4

When r= -4/3

==> a1= 3/4

a2= 3/4* (-4/3) = -1

a3= 3/4*(-4/3)^2 = 4/3

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If the 3 numbers are the consecutive terms of a geometric progression, we'll note them as:

a/r , a , a*r

From enunciation, we know that the product of the numbers is:

(a/r) * (a) * (a*r) = -1

We'll eliminate like terms:

a*a*a = -1

a^3 = -1 => a = -1

So, the numbers are:

-1/r , -1 , -1*r

Also from enunciation, we know that:

(-1/r) +( -1) + (-1*r) = 13r/12

To add the 3 numbers, we'll calculate LCD,which is r:

-1 -r - r^2 = 13r/12

-12 - 12r - 12r^2 = 13r

We'll move all terms to one side:

12r^2 + 12r + 12 + 13r = 0

12r^2 + 25r + 12 = 0

r1 = [-25+sqrt(625-576)]/24

r1 = (-25+7)/24

r1 = -18/24

r1 = -3/4

r2 = (-25-7)/24

r2 = -32/24

r2 = -4/3

So, when the common ratio is r = -3/4, the geometric progression is:

4/3 , -1 , 3/4

When the common ratio is r = -4/3, the G.P. is:

3/4 , -1 , 4/3