Arithmetic Series Worded

[mattgad]

A polygon has 16 sides. The lengths form a difference of d. The longest side is 6cm. And the perimeter of the polygon is 72cm.

Find the length of the shortest side.

I have so far figured out that:

n = 16
Sn = 72

a and d are both unknown.

U16 = 6
U16 = a + 15d

So,

6 = a + 15d
a = 6 - 15d

I'm not sure where to go next. Thanks.

 

[stapel]

I think there may be some information omitted in the statement of this exercise. Working from your subject line, I think it is supposed to read, in part, "The lengths of the polygon's sides for an arithmetic series with a common difference of d."

If so, then "n", if defined to be the number of sides, is equal to 16, and "Sn", if defined to be S_n, the sum of the related arithmetic series, is equal to 72. If "U16" means "a₁₆", the sixteenth term of the series, then it does equal 16 = a₁ + 15d.

Don't you have a formula for the sum of an arithmetic series? Something along the lines of:

. . . . .\(\displaystyle \Large{S_n\mbox{ }=\mbox{ }\frac{n}{2}\mbox{ }\left(a_1\mbox{ }+\mbox{ }a_n\right)}\)

Use this to create a second equation in terms of "a₁" (since you already know the values of n and a_n = a₁₆). Solve for the values of a₁.

If I have misinterpreted your post, please forgive my confusion and reply with clarification. Thank you.

Eliz.

 

[mattgad]

Yes, you are correct. I typed it wrong.

I've suddenly had an idea to use this formula.

\(\displaystyle Sn = n/2(2a + l)\)

Which is what I've been taught.

a being the first term, and l being last term.

So,

72 = 8(2a + 6)
72 = 48 + 16a
16a = 24
a = 24/16 = 1.5

Am I correct?

 

[mattgad]

My mistake, just looked at my notes, the formula is

Sn = n/2(a + l)

Meaning

a = 3