A circle is completely divided into n sectors in such a way that the a

1.	A circle is completely divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. If the smallest-of these angles is 8° and the largest 72°, calculate n and the angle in the fourth sector.
Answer» Let the common difference of the A.P. be x Given: The smallest angle = 8° ⇒ a = 8 And the largest is 72° ⇒ a_n = 72 ⇒ a + (n – 1)d = 72 ⇒ 8 + (n – 1)d = 72 ⇒ (n – 1) d = 72 – 8 = 64 ..........(1) We know that sum of all the angles of a circle is 360° Sn = n/2[(2a + (n-1)d] Putting the value of n in equation (1) we get (9 – 1) d = 64 d = 8 Now angle in fourth sector = a₄ = a + (4 – 1) d = a + 3d = 8 + 3 × 8 = 8 + 24 = 32 ∴ The value of n = 9 and angle in fourth sector is 32°

A circle is completely divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. If the smallest-of these angles is 8° and the largest 72°, calculate n and the angle in the fourth sector.

Answer»

Let the common difference of the A.P. be x
Given: The smallest angle = 8°
⇒ a = 8
And the largest is 72°
⇒ a_n = 72
⇒ a + (n – 1)d = 72
⇒ 8 + (n – 1)d = 72
⇒ (n – 1) d = 72 – 8 = 64 ..........(1)
We know that sum of all the angles of a circle is 360° Sn = n/2[(2a + (n-1)d]

Putting the value of n in equation (1) we get
(9 – 1) d = 64
d = 8
Now angle in fourth sector = a₄ = a + (4 – 1) d
= a + 3d = 8 + 3 × 8 = 8 + 24 = 32
∴ The value of n = 9 and angle in fourth sector is 32°

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A circle is completely divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. If the smallest-of these angles is 8° and the largest 72°, calculate n and the angle in the fourth sector.

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