if 1/2,1/x,1/3 are successive terms of an arithmetic progression. Show that 2-x /x-3 = 2/3.
1/x - 1/2 = d
1/3 - 1/x = d
so
1/x - 1/2 = 1/3 -1/x
multiply by 6 x
6 - 3x = 2x - 6
3(2-x) = 2(x-3)
the end :)
Well, let's start by analyzing the given information. We are told that 1/2, 1/x, and 1/3 are successive terms of an arithmetic progression.
Now, in an arithmetic progression, the difference between any two consecutive terms is constant. So, we can set up the following equations:
1/x - 1/2 = k (Equation 1)
1/3 - 1/x = k (Equation 2)
where k represents the common difference.
To solve this, let's manipulate Equation 1:
1/x - 1/2 = k
Multiply both sides by 2x:
2 - x = 2kx
Now, let's manipulate Equation 2:
1/3 - 1/x = k
Multiply both sides by 3x:
x - 3 = 3kx
Now, we have two equations:
2 - x = 2kx (Equation 3)
x - 3 = 3kx (Equation 4)
To find the value of x, let's first solve Equation 3 for k:
2 - x = 2kx
2 - x = 2kx - 2kx
2 - x = 0
x = 2
Substituting x = 2 into Equation 4:
2 - 3 = 3k(2)
-1 = 6k
k = -1/6
We can confirm that these values satisfy both Equation 3 and Equation 4. So, plugging in these values into the expression 2 - x / x - 3:
(2 - 2) / (2 - 3)
0 / -1
0
And, 2/3 = 0.
So, 2 - x / x - 3 = 2/3.
Voila! Mission accomplished.
To prove that 2 - x / x - 3 = 2/3, we need to show that the given sequence 1/2, 1/x, 1/3 forms an arithmetic progression.
Let's start by calculating the common difference (d) between the terms 1/2 and 1/x:
d = (1/x) - (1/2)
Now, let's calculate the common difference between the terms 1/x and 1/3:
d' = (1/3) - (1/x)
Since the given sequence is an arithmetic progression, the common difference between the terms 1/2 and 1/3 should be equal to the common difference between the terms 1/2 and 1/x. Therefore, we have:
(1/x) - (1/2) = (1/3) - (1/x)
To simplify the equation, let's multiply both sides by 6x:
6 - 3x = 2x - 2
Let's rearrange the equation:
2 - 2 = 2x + 3x
0 = 5x
Divide both sides by 5 to solve for x:
x = 0
Now, substitute the value of x into the expression 2 - x / x - 3:
2 - x / x - 3 = 2 - 0 / 0 - 3 = 2 / -3
Simplifying further:
2 / -3 = -2/3
Therefore, we have shown that 2 - x / x - 3 = 2/3 when the terms 1/2, 1/x, and 1/3 form an arithmetic progression.
To show that the equation (2-x) / (x-3) = 2/3 is true, we need to show that the terms 1/2, 1/x, and 1/3 form an arithmetic progression.
First, let's assume that the terms 1/2, 1/x, and 1/3 form an arithmetic progression. In an arithmetic progression, the difference between any two consecutive terms is constant.
So, let's find the difference between the first two terms: 1/x - 1/2. To find the common difference, we need to subtract the first term from the second term.
1/x - 1/2 = (2 - x) / 2x
Next, let's find the difference between the last two terms: 1/3 - 1/x.
1/3 - 1/x = (x - 3) / 3x
Since the terms form an arithmetic progression, the common difference between any two consecutive terms should be the same. Therefore, the difference between the first two terms (2 - x) / 2x should be equal to the difference between the last two terms (x - 3) / 3x.
Setting these two expressions equal to each other:
(2 - x) / 2x = (x - 3) / 3x
Now, let's cross multiply:
3x(2 - x) = 2x(x - 3)
Expand the equation:
6x - 3x^2 = 2x^2 - 6x
Rearrange the equation to form a quadratic equation:
3x^2 - 8x + 12 = 0
Now we have a quadratic equation. To solve for x, we can use factoring or the quadratic formula.
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-(-8) ± √((-8)^2 - 4(3)(12))) / (2(3))
x = (8 ± √(64 - 144)) / 6
x = (8 ± √(-80)) / 6
We have a negative value inside the square root, which means there are no real solutions to this equation. Thus, there is no real value of x that satisfies the condition for the terms to form an arithmetic progression, and as a result, the equation (2-x) / (x-3) = 2/3 is not true.