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#