Advertisements
Advertisements
Question
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.
Solution
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.
APPEARS IN
RELATED QUESTIONS
If (a + b + c) (a - b + c) = a2 + b2 + c2 show that a, b, c are in continued proportion.
If `a/b = b/c` and `a , b , c > 0` then show that, (a + b + c) (b - c) = ab - c2
If `a/b = b/c` and a, b, c > 0 then show that, `[a^2 + b^2]/(ab) = (a+c)/b`
If a, b, c are in continued proportion, then prove that
`a/[ a + 2b] = [a - 2b]/[ a - 4c]`
If a, b, c are in continued proportion, then prove that
`b/[b+c] = [a-b]/[a-c]`
If a, b, c, d are in proportion, then prove that
`[11a^2 + 9ac]/[ 11b^2 + 9bd] = [ a^2 + 3ac]/[ b^2 + 3bd]`
If a, b, c, d are in proportion, then prove that
`sqrt[( a^2 + 5c^2)/( b^2 + 5d^2)] = a/b`
If a, b, c, d are in proportion, then prove that
`[a^2 + ab + b^2]/[a^2 - ab + b^2] = [c^2 + cd + d^2 ]/[ c^2 - cd + d^2 ]`
