If the product of three numbers in geometric progression is 216 and their sum is 19.?

Question

find the largest of the three numbers. I had to continue quesiton in here...

Answer 1

If a is the middle term of the progression, the others are a/q and a*q, for some q>0. Since the product of such terms is a^3, we have a^3 = 216 => a = 6.

In addition, we have 6/q + 6 +6q = 19 => 6/q + 6*q = 13 => 6*q^2 -13*q + 6 = 0 Ths is a quadratic equation, whose solutions, applying Bhaskara, are q1 = 3/2 and q2 = 2/3. So, according to your choice, the terms of the progression are 4, 6 , 9 or 9, 6 , 4. In any case, the largest number is 9.

Answer 2

The numbers in the geometric progression have the form

a, ar, ar^2

for some a and r.

Their product is a^3r^3 = 216, and
their sum is a+ar+ar^2=a(1+r+r^2)=19.

Since (ar)^3=216=6^3, it follows that ar=6.

Therefore, 19=a+ar+ar^2=a+6+ar^2, so

a+ar^2=a+ar*r=a+6r=19-6=13.

We now have to solve a+6r=13, but r=6/a, so now we have a+36/a=13. Multiply by a:

a^2+36=13a, or a^2-13a+36=0.

Factor: (a-4)(a-9)=0, so a=4 or a=9.

If a=4, then r=6/a=3/2. Otherwise, a=9, and r=2/3.

Either way, the numbers in the geometric progression are 4, 6 and 9, so the largest is 9.

Answer 3

9

a(ar)(ar^2) = 216
(ar)^3 = 216
ar = 6

a + ar + ar^2 = 19
a + 6 + ar^2 = 19
ar^2 = 13-a

a(ar)(ar^2) = 216
a(6)(13-a) = 216
13a - a^2 = 36
a^2 -13 a + 36 = 0
(a-9)(a-4)=0
a= 9 or a = 4
if a=9, r= 2/3
if a=4, r=3/2

test
4 + 4(3/2) + 4(3/2)^2 =
4 + 6 + 9 = 19

9 + 9(2/3) + 9(2/3)^2 =
9 + 6 + 4 = 19