Q. 44.0( 4 Votes )

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

Answer :

Let us label the diagram

Let the distance between two consecutive vertical lines be 'x' and height of vertical lines be h_{1}, h_{2}, h_{3}, …, and so on.

Let AB_{1} = x

Now, we know

⇒ perpendicular = base × tanθ

Therefore, in consecutive triangles, we have

B_{1}C_{1} = AB_{1} × tan 40°

⇒ h_{1} = atan40°

B_{2}C_{2} = AB_{2} × tan 40°

⇒ h_{2} = (a + x)tan40°

B_{3}C_{3} = AB_{3} × tan 40°

⇒ h_{3} = (a + 2x)tan40°

.

.

.

and so on.

Now, we have

h_{1}, h_{2}, h_{3},… = atan40°, (a + x)tan40°, (a + x)tan40°, …

Let us calculate the common difference.

h_{2} - h_{1} = (a + x)tan40° - atan40° = xtan40°

h_{3} - h_{2} = (a + 2x)tan40° - (a + x)tan40° = xtan40°

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

h_{1}, h_{2}, h_{3}, …. 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]

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