Finding the Number of Terms and Common Difference in a Finite Series

[jade35]

I'm in Algebra 2, 8th grade. This question is driving me nuts! My book and notes do not help at all.

The sum of a series is 2125. The first term is 43 and the last term is 127. How many terms are there, and what is the common difference?

I have no idea how to find the terms, because all of the equations I know have d in there.. and I don't know if I'm supposed to find the D first, or whatever.

 

[berkeman]

What is the basic property of an arithmetic series? A constand delta between each term, right? That's probably the d you are referring to.

So this series is (43 + 0) + (43 + d) + (43 + 2d) + ... + 127 = 2125.

How many 43's does it take to make 2125? Then the number of terms is less than that number. Given the number of terms n, how many 43's are there, and how many d's. Does that help guide you to the answer?

 

[jade35]

A little bit. So I just keep going (43 + 3d) and so on and so on? How will I finally find what d equals?

 

[berkeman]

Write the equation for the sum in terms of d and n. Then your solutions for n have to be a whole number, although I suppose d does not have to be whole. If you get multiple solutions for non-whole d and whole n, I'd pick the answer with a whole number for both if it exists.

 

[jade35]

The a(n)= a1 + (n-1)d equation?

 

[berkeman]

No no no. Like this:

n=3: (43+0) + (43+d) + (43+2d) = 129 + 2d = 2125
n=4: (43+0) + (43+d) + (43+2d) + (43+3d) = 172 + 6d = 2125
n=...

general n: <<write the equation>>

Then solve for several n and d to see what looks reasonable...

 

[berkeman]

Gotta go. Good luck!

 

[jade35]

thank you! I'm a wee bit closer now.

 

[berkeman]

thank you! I'm a wee bit closer now.

You're welcome. I'm just checking in from home now briefly. BTW, you will have a formula for the finite series of the sum of the multiple d terms in terms of n. That will factor into the final equations.