If the continued product of three numbes in g.p. is 216 and the sum of their products in pair is 156. find the numbers.
apparently three numbers are in a GP
then the 3 numbers are: a, ar, and ar^2
"the continued product of three numbes in g.p. is 216"
I am guessing you mean:
a(ar)(ar^2) = 216
a^3 r^3 = 216
ar = 6
a = 6/r
"the sum of their products in pair is 156"
a(ar) + a(ar^2) + ar(ar^2) = 156 ???
a^2 r + a^2 r^2 + a^2 r^3) = 156
(ar)(a + ar + ar^2) = 156
6a(1 + r + r^2) = 156
a(1+r+r^2) = 26
(6/r)(1 + r + r^2) = 26
6/r + 6 + 6r = 26
times r
6 + 6r + 6r^2 = 26r
6r^2 - 20r + 6 = 0
(r - 3)(6r - 2) = 0
r = 3 or r = 1/3
carry on
To solve this problem, let's call the three numbers in geometric progression (g.p.) x, y, and z.
The first condition tells us that the continued product of the three numbers is 216, which means:
x * y * z = 216 ............. (Equation 1)
The second condition states that the sum of their products taken in pairs is 156. This means:
xy + yz + xz = 156 ............ (Equation 2)
We now have two equations with two unknowns, so we can solve them simultaneously.
From Equation 1, we can observe that the product of the three numbers is 216. Hence,
x * y * z = 216
We need to find three numbers whose product is 216. We can start by trying different sets of numbers and checking if their product matches 216.
One possible approach is to start with a trial and error method. We can use factors of 216 to find the numbers.
The factors of 216 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, and 216.
Now, we need to select three numbers from this list such that they form a geometric progression.
Let's try selecting the numbers 4, 6, and 9.
Calculating their product:
4 * 6 * 9 = 216
This matches the first condition. Now, let's substitute these values into Equation 2 to see if they satisfy the second condition:
4*6 + 6*9 + 4*9 = 24 + 54 + 36 = 114
Since 114 is not equal to 156, this set of numbers does not satisfy the second condition.
We need to continue trying different combinations of factors until we find a set of numbers that satisfy both conditions.
Let's try another set of numbers from the factors of 216: 2, 6, and 18.
Calculating their product:
2 * 6 * 18 = 216
This satisfies the first condition. Now, let's substitute these values into Equation 2:
2*6 + 6*18 + 2*18 = 12 + 108 + 36 = 156
Since 156 is equal to 156, this set of numbers satisfies the second condition.
Therefore, the three numbers in geometric progression such that their continued product is 216 and the sum of their products in pair is 156 are 2, 6, and 18.