A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is $8^{0}$ and the largest is $72^{0}$ , calculate n and the angle in the fifth sector.

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Solution

Given,

$a = 8$

$a_{n} = 72$

$S_{n} = \frac{n}{2} [a + l]$

$360 = \frac{n}{2} [8 + 72]$

$n = \frac{360 \times 2}{8 + 72} = 9$

$a_{n} = a + (n - 1) d$

$72 = 8 + (9 - 1) d$

$d = \frac{72 - 8}{8} = 8$

$a_{5} = a + (5 - 1) d$

$= 8 + 4 \times 8 = 40$

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A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is 80 and the largest is 720, calculate n and the angle in the fifth sector.

Given,a=8an=72Sn=n2[a+l]360=n2[8+72]n=360×28+72=9an=a+(n−1)d72=8+(9−1)dd=72−88=8a5=a+(5−1)d=8+4×8=40

A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is $8^{0}$ and the largest is $72^{0}$ , calculate n and the angle in the fifth sector.

Given,

$a = 8$

$a_{n} = 72$

$S_{n} = \frac{n}{2} [a + l]$

$360 = \frac{n}{2} [8 + 72]$

$n = \frac{360 \times 2}{8 + 72} = 9$

$a_{n} = a + (n - 1) d$

$72 = 8 + (9 - 1) d$

$d = \frac{72 - 8}{8} = 8$

$a_{5} = a + (5 - 1) d$

$= 8 + 4 \times 8 = 40$