In the picture, the vertical lines are equally spaced. Prove that their heights are in arithmetic sequence. What is the common difference?

Let us label the diagram

Let the distance between two consecutive vertical lines be 'x' and height of vertical lines be h₁, h₂, h₃, …, and so on.

Let AB₁ = x

Now, we know

⇒ perpendicular = base × tanθ

Therefore, in consecutive triangles, we have

B₁C₁ = AB₁ × tan 40°

⇒ h₁ = atan40°

B₂C₂ = AB₂ × tan 40°

⇒ h₂ = (a + x)tan40°

B₃C₃ = AB₃ × tan 40°

⇒ h₃ = (a + 2x)tan40°

.

.

.

and so on.

Now, we have

h₁, h₂, h₃,… = atan40°, (a + x)tan40°, (a + x)tan40°, …

Let us calculate the common difference.

h₂ - h₁ = (a + x)tan40° - atan40° = xtan40°

h₃ - h₂ = (a + 2x)tan40° - (a + x)tan40° = xtan40°

As, the common difference between the terms is same, terms are in AP and hence,

h₁, h₂, h₃, …. are in AP.

Therefore, heights of vertical lines are in AP with common difference xtan40°, where 'x' is the space between two consecutive lines. [tan 40° = 0.8391]