Question

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.​

Answer 1

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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