The angles in an n-sided polygon satisfy an arithmetic sequence with a_1 =143^@ and d=2. Given that all of the internal angles of the polygon are less than 180^@, how do I find n?

May 21, 2017

Using a_1 = 143°

There are 18 sides, $n = 18$

Explanation:

Think about the size of the angles.
If a = 143° and d=2, the sequence is:

143°," "145°," "147°," "149° and so on

${T}_{n} = a + \left(n - 1\right) d \text{ } \leftarrow$ with $a = 143 \mathmr{and} d = 2$

${T}_{n} = 143 + 2 n - 2$

${T}_{n} = 141 + 2 n$

The last term $\left({T}_{n}\right)$ must still be less than $180$

$180 > 141 + 2 n$
$39 > 2 n$
$19.5 > n$

$n = 19$

So there may be $19$ sides.

We can check this by finding the sum of the first $19$ terms and the answer should be the same as the sum of the interior angles:

180(n-2) = 180 xx17 = 3060°

${S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]$

${S}_{19} = \frac{19}{2} \left[2 \left(143\right) + \left(18\right) 2\right]$

$= \frac{19}{2} \times 322$

=3059°

Adding an odd number of odd angles ($19$) will never give an even number.

The sum is not 3060°, so 19 sides does not work.

Let's check with $18$ sides:

S_18 = 18/2[2(143) +(17)2] = 2880°

Sum interior angles = $180 \times 16 = 2880$

Thus $n = 18$