If sum of three numbers in G.P. is 38, and their product is 1728, find the numbers.
Answers
Answered by
155
let the numbers are a/r, a and ar
product = 1728
⇒ a/r * a * ar = 1728
⇒ a³ = 1728
⇒ a = 12
12/r + 12 + 12r = 38
⇒ 12 + 12r + 12r² = 38r
⇒ 12r² - 26r + 12 = 0
⇒ 6r² - 13r + 6 = 0
r = 2/3 and 3/2
numbers are 8, 12 and 18
product = 1728
⇒ a/r * a * ar = 1728
⇒ a³ = 1728
⇒ a = 12
12/r + 12 + 12r = 38
⇒ 12 + 12r + 12r² = 38r
⇒ 12r² - 26r + 12 = 0
⇒ 6r² - 13r + 6 = 0
r = 2/3 and 3/2
numbers are 8, 12 and 18
Answered by
38
a + ar + ar² = 38
a(1+r+r²) = 38 ----(1)
(a)(ar)ar² = 1728
a³r³ = 1728
(ar)³ = 1728
ar = 12
➺ r = 12/a
putting this value in equation (1)
a( 1+12/a + 144/a²) = 38
➺ a² + 12a + 144 = 38a
➺ a²-26a + 144 = 0
➺ (a-18)(a-8) = 0
➺ a = 18 & a = 8
➺ r = 2/3 & r = 3/2
r = 12/a
putting this we get
➺ a(1+12/a + 144/a²) = 38
➺ a²-26a + 144 = 0
➺(a-18)(a-8) = 0
➺ a = 18 & a = 8
➺ r = 2/3 & r = 3/2
So the numbers are 18 , 12 & 8
a(1+r+r²) = 38 ----(1)
(a)(ar)ar² = 1728
a³r³ = 1728
(ar)³ = 1728
ar = 12
➺ r = 12/a
putting this value in equation (1)
a( 1+12/a + 144/a²) = 38
➺ a² + 12a + 144 = 38a
➺ a²-26a + 144 = 0
➺ (a-18)(a-8) = 0
➺ a = 18 & a = 8
➺ r = 2/3 & r = 3/2
r = 12/a
putting this we get
➺ a(1+12/a + 144/a²) = 38
➺ a²-26a + 144 = 0
➺(a-18)(a-8) = 0
➺ a = 18 & a = 8
➺ r = 2/3 & r = 3/2
So the numbers are 18 , 12 & 8
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