SATHEE: Sequences and Series 2 Question 19

Sequences and Series 2 Question 19

20.

The interior angles of a polygon are in arithmetic progression. The smallest angle is $120^{\circ}$ and the common difference is $5^{\circ}$ . Find the number of sides of the polygon.

(1980, 3M)

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Answer:

Correct Answer: 20. (9)

Solution:

Since, angles of polygon are in an AP.

$∴$ Sum of all angles

$= (n - 2) \times 180^{\circ} = \frac{n}{2} 2 (120^{\circ}) + (n - 1) 5^{\circ}$

$\begin{aligned} \Rightarrow 5 n^{2} - 125 n + 720 = 0 \\ \Rightarrow n^{2} - 25 n + 144 = 0 \\ \Rightarrow (n - 9) (n - 16) = 0 \\ \Rightarrow n = 9, 16 \end{aligned}$

If $n = 9$ , then largest angle $= a + 8 d = 160^{\circ}$

Again, if $n = 16$ , the $n$ largest angle

$= a + 15 d = 120^{\circ} + 75 = 195^{\circ}$

which is not possible.

[since, any angle of polygon cannot be $> 180^{\circ}$ ]

Hence, $n = 9$

[neglecting $n = 16$ ]

Sequences and Series 2 Question 19

20.

Answer:

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Sequences and Series 2 Question 19

20.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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