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If (1)/(a)+(1)/(c )=(1)/(2b-a)+(1)/(2b-c...

If `(1)/(a)+(1)/(c )=(1)/(2b-a)+(1)/(2b-c)`, then

A

`a,b,c` are in `A.P.`

B

`a,(b)/(2),c` are in `A.P.`

C

`a,(b)/(2),c` are in `H.P.`

D

`a,2b,c` are in `H.P.`

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To solve the equation \[ \frac{1}{a} + \frac{1}{c} = \frac{1}{2b-a} + \frac{1}{2b-c}, \] we will first manipulate the equation to simplify it. ### Step 1: Rewrite the equation We start with the given equation: \[ \frac{1}{a} + \frac{1}{c} = \frac{1}{2b-a} + \frac{1}{2b-c}. \] ### Step 2: Combine the fractions on both sides We can combine the fractions on both sides of the equation. The left-hand side becomes: \[ \frac{c + a}{ac}. \] The right-hand side becomes: \[ \frac{(2b-c) + (2b-a)}{(2b-a)(2b-c)} = \frac{4b - (a+c)}{(2b-a)(2b-c)}. \] So we rewrite the equation as: \[ \frac{c + a}{ac} = \frac{4b - (a+c)}{(2b-a)(2b-c)}. \] ### Step 3: Cross-multiply Cross-multiplying gives us: \[ (c + a)(2b - a)(2b - c) = ac(4b - (a + c)). \] ### Step 4: Expand both sides Now we expand both sides. The left-hand side expands to: \[ (c + a)(2b - a)(2b - c) = (c + a)(4b^2 - 2bc - 2ab + ac). \] The right-hand side expands to: \[ ac(4b - a - c). \] ### Step 5: Set the equation to zero We can rearrange the equation to set it to zero: \[ (c + a)(4b^2 - 2bc - 2ab + ac) - ac(4b - a - c) = 0. \] ### Step 6: Analyze the equation From the rearranged equation, we can analyze the terms. The equation implies that either: 1. \( a + c - 2b = 0 \) (which means \( a, b, c \) are in arithmetic progression), or 2. The entire expression equals zero. ### Conclusion Thus, we conclude that \( a + c = 2b \), indicating that \( a, b, c \) are in arithmetic progression.
To solve the equation \[ \frac{1}{a} + \frac{1}{c} = \frac{1}{2b-a} + \frac{1}{2b-c}, \] we will first manipulate the equation to simplify it. ...
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