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प्रश्न
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
उत्तर
Let the remaining two numbers be x and y.
So, the numbers x, 12, y are in continued proportion.
∴ xy = (12)2 = 144 ...(1)
Also,
x + y = 26 ...(2)
Solving (1) and (2), we get
`x + 144/x = 26`
⇒ `(x^2+144)/x=26`
⇒ `x^2 + 144 = 26x`
⇒ `x^2 - 26x +144 = 0`
⇒ `x^2 - 18x - 8x + 144 = 0`
⇒ `x( x - 18) - 8( x - 18) = 0`
⇒ `( x - 18 )( x - 8 )=0`
⇒ `x - 8 = 0 or x - 18 = 0`
⇒ `x = 8 or x = 18`
when x = 8, y = 26 - 8 = 18 ...[using 2]
when x = 18, y = 26 - 18 = 8
Thus, the numbers are 8, 12, 18 or 18, 12, 8.
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