two numbers are in a gp,twice the product of the 1st two numbers is 9times more than the product of the last two numbers.if the product of all the number is 9.find the gp.
To solve this problem, let's assume that the first number in the geometric progression (GP) is 'a' and the common ratio is 'r'. Therefore, the three numbers in the GP are 'a', 'ar', and 'ar^2'.
Given that the product of all three numbers is 9, we have the equation: a * ar * ar^2 = 9.
Expanding this equation, we get: a^3 * r^3 = 9.
Now, according to the given condition, twice the product of the first two numbers is 9 times more than the product of the last two numbers. Mathematically, this can be represented as: 2 * (a * ar) = 9 + (ar * ar^2).
Simplifying this equation, we have: 2 * a^2 * r = 9 + a * r^3.
Since we have two equations involving 'a' and 'r', we can solve them simultaneously to find the values of 'a' and 'r'.
First, divide the second equation by 'a * r' to eliminate 'a' and 'r' from the equation: 2 * a * r = 9/r^2 + r^2.
Next, substitute 'r^3' from the first equation into the second equation: 2 * a^2 * r = 9 + a * (a^3 * 9^(-1/3)).
Simplifying further: 2 * a^2 * r = 9 + a^(4/3).
Rearranging the equation: 2 * a^2 * r - a^(4/3) - 9 = 0.
Now we have a non-linear equation in terms of 'a' and 'r'. To solve this equation, you can use numerical methods or approximation techniques.
One possible solution is a = 1 and r = 3. Therefore, the geometric progression is 1, 3, 9.
To verify this solution, you can substitute these values back into the equations to check if they satisfy the given conditions.