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 $n$ ?

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Answer 1 · 2017-05-21T13:59:34

Using $a_1 = 143°$

There are 18 sides, $n = 18$

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

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

$T_n = a +(n-1)d" "larr$ with $a = 143 and d =2$

$T_n = 143+2n-2$

$T_n = 141 +2n$

The last term $(T_n)$ must still be less than $180$

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

$n=19$

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:

$180(n-2) = 180 xx17 = 3060°$

$S_n = n/2[2a +(n-1)d]$

$S_19 = 19/2[2(143) +(18)2]$

$= 19/2 xx 322$

$=3059°$

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 = $180 xx16 =2880$

Thus $n=18$

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