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Three numbers are in the ratio of 1:2:3. If 2, 4, and 11 are added to them respectively, the resulting numbers are in G.P. Find the original three numbers.

Sagot :

To find the original three numbers in the ratio 1:2:3, we can follow these steps:

1. Define the original numbers: Let the original numbers be [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]. Since they are in the ratio 1:2:3, we can represent them as:
[tex]\[ a = x, \quad b = 2x, \quad c = 3x \][/tex]
where [tex]\( x \)[/tex] is a positive real number.

2. Add the given values to each number:
- To [tex]\( a \)[/tex], add 2 => [tex]\( a + 2 \)[/tex]
- To [tex]\( b \)[/tex], add 4 => [tex]\( b + 4 \)[/tex]
- To [tex]\( c \)[/tex], add 11 => [tex]\( c + 11 \)[/tex]

3. Resulting numbers in G.P.: The resulting numbers should be in Geometric Progression (G.P.). Therefore:
- The three numbers in G.P. are [tex]\( (a + 2) \)[/tex], [tex]\( (b + 4) \)[/tex], and [tex]\( (c + 11) \)[/tex].
- In a G.P., the square of the middle term is equal to the product of the first and the third terms:
[tex]\[ (b + 4)^2 = (a + 2)(c + 11) \][/tex]

4. Substitute [tex]\( a = x \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = 3x \)[/tex] into the equation:
[tex]\[ (2x + 4)^2 = (x + 2)(3x + 11) \][/tex]

5. Expand and solve the equation:
[tex]\[ (2x + 4)^2 = (x + 2)(3x + 11) \][/tex]
[tex]\[ 4x^2 + 16x + 16 = 3x^2 + 11x + 6x + 22 \][/tex]
[tex]\[ 4x^2 + 16x + 16 = 3x^2 + 17x + 22 \][/tex]

6. Simplify by moving all terms to one side of the equation:
[tex]\[ 4x^2 + 16x + 16 - 3x^2 - 17x - 22 = 0 \][/tex]
[tex]\[ x^2 - x - 6 = 0 \][/tex]

7. Solve the quadratic equation [tex]\( x^2 - x - 6 = 0 \)[/tex]:
- This can be factored into:
[tex]\[ (x - 3)(x + 2) = 0 \][/tex]
- Hence, the solutions are:
[tex]\[ x = 3 \quad \text{or} \quad x = -2 \][/tex]
Since [tex]\( x \)[/tex] represents a positive quantity (as it denotes the magnitude of ratio), we discard [tex]\( x = -2 \)[/tex] and accept [tex]\( x = 3 \)[/tex].

8. Find the original numbers:
[tex]\[ a = x = 3, \quad b = 2x = 6, \quad c = 3x = 9 \][/tex]

So, the original three numbers are:
[tex]\[ \boxed{3, 6, 9} \][/tex]
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