If the product of three consecutive numbers in G.P. is 216 and the sum of the product taken in pairs is 156, find the numbers.
Solution:- Let the first three terms are a/r ,a, a * r We are given , Product of three terms = 216 So , (a/r) x a x a * r = 216 a^3 = 6^3
a = 6
Now, sum of their product in pairs = 156
(a/r) x a + (a) x (ar) + (a r) x (a/r) = 156
a²/r + a² r + a² = 156
a²[(1/r) + r + 1] = 156
a² [ (1 + r² + r)/r] = 156
a² (r²+r+1)/r = 156
(6²/r)(r²+r+1) = 156
(r²+r+1)/r = 156/36
(r²+r+1)/r = 13/3
3 (r²+r+1) = 13 r
3 r² + 3 r + 3 - 13 r = 0
3 r² - 10 r + 3 = 0 Now , we will find the factors , so we get (3r -1) (r-3) = 0 We will equate with zero ( 3r-1 ) = 0 , ( r - 3) = 0 So , 3r = 1 , r = 3 or r = 1/3 and r = 3 Now we will plug the values of a and r ( We got a = 6 and r = 1/3 , 3) So first three terms was a/ r , a , a* r First we will plug a = 6 and r = 3 So a/ r = 6 / 3 => 2 a = 6 a * r = 6 * 3 = >18 Now we will plug a = 6 and r = 1/3 So a/ r = 6 / (1/3 ) => 18 a = 6 a * r = 6 * 1/ 3 => 2 Therefore the three terms are 18,6,2 or 2,6,18 That would be the final answer.
a^3 = 6^3
a = 6
Now, sum of their product in pairs = 156
(a/r) x a + (a) x (ar) + (a r) x (a/r) = 156
a²/r + a² r + a² = 156
a²[(1/r) + r + 1] = 156
a² [ (1 + r² + r)/r] = 156
a² (r²+r+1)/r = 156
(6²/r)(r²+r+1) = 156
(r²+r+1)/r = 156/36
(r²+r+1)/r = 13/3
3 (r²+r+1) = 13 r
3 r² + 3 r + 3 - 13 r = 0
3 r² - 10 r + 3 = 0 Now , we will find the factors , so we get (3r -1) (r-3) = 0 We will equate with zero ( 3r-1 ) = 0 , ( r - 3) = 0 So , 3r = 1 , r = 3 or r = 1/3 and r = 3 Now we will plug the values of a and r ( We got a = 6 and r = 1/3 , 3) So first three terms was a/ r , a , a* r First we will plug a = 6 and r = 3 So a/ r = 6 / 3 => 2 a = 6 a * r = 6 * 3 = >18 Now we will plug a = 6 and r = 1/3 So a/ r = 6 / (1/3 ) => 18 a = 6 a * r = 6 * 1/ 3 => 2 Therefore the three terms are 18,6,2 or 2,6,18 That would be the final answer.
