In this picture, the perpendiculars
to the bottom line are equally
spaced. Prove that, continuing like
this, the lengths of perpendiculars
form an arithmetic sequence.
Answers
Given : the perpendiculars to the bottom line are equally spaced.
To find : Prove that, continuing like this, the lengths of perpendiculars
form an arithmetic sequence.
Solution:
Let say Horizontal Distance of First Perpendicular = a
and then Equally Spaced at d
Hence Distances area
a , a + d , a + 2d , .................
Lets compare two triangle one with base a and another with a + d
Now one angle is common
& another is 90°
Hence Similar Triangle
=> (a + d)/a = p₂/p₁
=> p₂/p₁ = 1 + d/a
=> p₂ = p₁ + p₁ (d/a)
=> p₂ = p₁ + (p₁d/a)
(a + 2d)/a = p₃/p₁
=> 1 + 2d/a = p₃/p₁
=> p₃ = p₁ + (p₁/a)2d
=> p₃ = p₁ + 2(p₁d/a)
(a + (n-1)d)/a = pₙ/p₁
=> 1 + (n-1)d/a = pₙ/p₁
=> pₙ = p₁ + (n-1)(p₁d/a)
p₁ , p₁ + (p₁d/a) , p₁ + 2(p₁d/a) , .................... p₁ + (n-1)(p₁d/a)
This is an AP
Where first term = p₁
& common difference = p₁d/a
Hence proved lengths of perpendiculars form an arithmetic sequence.
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