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If a,b,c are in AP , x is the GM between a and b , y is the GM between b and c ,show that b^2 is the AM between x^2 and y^2?
Verified Answer
If a,b,c are in AP , x is the GM between a and b , y is the GM between...
Since a,b,c are in AP so b is the mean of a and c
b = (a+c)/2 ⇒ a+c = 2b ---(1)
Now , x is GM of a and b
x = √(ab)
x2 = ab
And y is GM of b and c
y2 = bc
Now x2 + y2 = ab +bc = b(a+c) = 2b2 (from eq (1))
⇒ b2= (x2 + y2)/2
or b2 is the mean of (x2 + y2)
Thus proved.
This question is part of UPSC exam. View all Class 11 courses
Most Upvoted Answer
If a,b,c are in AP , x is the GM between a and b , y is the GM between...
Understanding Arithmetic Progression (AP)
When a, b, c are in Arithmetic Progression (AP), it follows the relation:
- b - a = c - b
This implies:
- 2b = a + c
Geometric Means (GM)
The geometric mean (GM) between two numbers a and b is given by:
- x = √(ab)
Similarly, the GM between b and c is:
- y = √(bc)
Finding x^2 and y^2
Calculating the squares of the GMs:
- x^2 = ab
- y^2 = bc
Proving b^2 as the AM between x^2 and y^2
To show that b^2 is the Arithmetic Mean (AM) of x^2 and y^2, we need to verify the following:
- AM of x^2 and y^2 = (x^2 + y^2) / 2
- We need to prove:
b^2 = (x^2 + y^2) / 2
Substituting the values of x^2 and y^2:
- (x^2 + y^2) = ab + bc
Thus:
- AM = (ab + bc) / 2
Now, substituting 2b = a + c into our equation:
- b = (a + c) / 2
Thus, we can rewrite it as:
- b^2 = (ab + bc) / 2
Conclusion
Hence, we have demonstrated that:
- b^2 is indeed the AM between x^2 and y^2, confirming the relationship in the context of AP.
When a, b, c are in Arithmetic Progression (AP), it follows the relation:
- b - a = c - b
This implies:
- 2b = a + c
Geometric Means (GM)
The geometric mean (GM) between two numbers a and b is given by:
- x = √(ab)
Similarly, the GM between b and c is:
- y = √(bc)
Finding x^2 and y^2
Calculating the squares of the GMs:
- x^2 = ab
- y^2 = bc
Proving b^2 as the AM between x^2 and y^2
To show that b^2 is the Arithmetic Mean (AM) of x^2 and y^2, we need to verify the following:
- AM of x^2 and y^2 = (x^2 + y^2) / 2
- We need to prove:
b^2 = (x^2 + y^2) / 2
Substituting the values of x^2 and y^2:
- (x^2 + y^2) = ab + bc
Thus:
- AM = (ab + bc) / 2
Now, substituting 2b = a + c into our equation:
- b = (a + c) / 2
Thus, we can rewrite it as:
- b^2 = (ab + bc) / 2
Conclusion
Hence, we have demonstrated that:
- b^2 is indeed the AM between x^2 and y^2, confirming the relationship in the context of AP.
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If a,b,c are in AP , x is the GM between a and b , y is the GM between b and c ,show that b^2 is the AM between x^2 and y^2?
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If a,b,c are in AP , x is the GM between a and b , y is the GM between b and c ,show that b^2 is the AM between x^2 and y^2? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about If a,b,c are in AP , x is the GM between a and b , y is the GM between b and c ,show that b^2 is the AM between x^2 and y^2? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a,b,c are in AP , x is the GM between a and b , y is the GM between b and c ,show that b^2 is the AM between x^2 and y^2?.
If a,b,c are in AP , x is the GM between a and b , y is the GM between b and c ,show that b^2 is the AM between x^2 and y^2? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about If a,b,c are in AP , x is the GM between a and b , y is the GM between b and c ,show that b^2 is the AM between x^2 and y^2? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a,b,c are in AP , x is the GM between a and b , y is the GM between b and c ,show that b^2 is the AM between x^2 and y^2?.
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