show that (a-b)^2,(a^2+b^2) and( a+b)^2 are in AP.
Answers
Answer- The above question is from the chapter 'Arithmetic Progressions'.
Concept used: a, b and c are terms of some sequence.
If b - a = c - b, then a, b and c are in AP.
In other words, for a sequence to be in AP, common difference (d) should be same throughout.
Given question: Show that (a - b)² , (a² + b²) and (a + b)² are in AP.
Solution: Let a = (a - b)²
a = a² + b² - 2ab
a₂ = (a² + b²)
a₃ = (a + b)²
a₃ = a² + b² + 2ab
a₂ - a = (a² + b²) - (a - b)²
a₂ - a = a² + b² - a² - b² + 2ab
a₂ - a = 2ab = d₁
a₃ - a₂ = (a + b)² - (a² + b²)
a₃ - a₂ = a² + b² + 2ab - a² - b²
a₃ - a₂ = 2ab = d₂
∵ d₁ = d₂ i.e. common difference is same throughout, ∴ (a - b)² , (a² + b²) and (a + b)² are in AP.
Answer:
Answer- The above question is from the chapter 'Arithmetic Progressions'.
Concept used: a, b and c are terms of some sequence.
If b - a = c - b, then a, b and c are in AP.
In other words, for a sequence to be in AP, common difference (d) should be same throughout.
Given question: Show that (a - b)² , (a² + b²) and (a + b)² are in AP.
Solution: Let a = (a - b)²
a = a² + b² - 2ab
a₂ = (a² + b²)
a₃ = (a + b)²
a₃ = a² + b² + 2ab
a₂ - a = (a² + b²) - (a - b)²
a₂ - a = a² + b² - a² - b² + 2ab
a₂ - a = 2ab = d₁
a₃ - a₂ = (a + b)² - (a² + b²)
a₃ - a₂ = a² + b² + 2ab - a² - b²
a₃ - a₂ = 2ab = d₂
∵ d₁ = d₂ i.e. common difference is same throughout, ∴ (a - b)² , (a² + b²) and (a + b)² are in AP.