Write a rule for the nth term of the arithmetic sequence. ( I'm not sure how I'm suppose to write the numbers below a :C so bare with me )
a(2) = - 28, a(20) = 52
I used the formula a(n) = a(1) + ( n-1)d
a(20) = a(1) + (20-1)d > 52 = a(1)+19d
a(2) = a(1) + (2-1)d > -28= a(1) + 1d
And that's where I'm stuck >_< please help. Thanks :)
just write down both equations, using you known values for a(2) and a(20):
a + 19d = 52
a + d = -28
subtract the equations, and the a goes away!
18d = 80
d = 40/9
so, a = -28 - 40/9 = -292/9
So, leaving out all the nasty nines, the numerators (multiplied by 9) proceed as follows:
-292 -252 -212 -172 -132
-92 -52 -12 28 68
108 148 188 228 268
308 348 388 428 468
Note that a(2) = -252/9 = -28
and a(20) = 468/9 = 52
To find the rule for the nth term of an arithmetic sequence, you need to determine the common difference (d) and the value of the first term (a(1)).
Given that:
a(2) = -28,
a(20) = 52.
We can use the information to set up two equations and solve them simultaneously to find the values of a(1) and d.
Using the formula a(n) = a(1) + (n-1)d:
For a(2), we have:
-28 = a(1) + 1d -- Equation 1
For a(20), we have:
52 = a(1) + 19d -- Equation 2
To solve for a(1) and d, we can subtract Equation 1 from Equation 2:
52 - (-28) = (a(1) + 19d) - (a(1) + 1d)
Simplifying, we get:
80 = 18d
Now, to find the value of d:
Divide both sides by 18:
80/18 = d
d ≈ 4.44 (rounded to two decimal places).
Now that you have the value of d, you can substitute it into either Equation 1 or Equation 2 to find the value of a(1).
Using Equation 1:
-28 = a(1) + 1(4.44)
Simplifying, we get:
-28 = a(1) + 4.44
To isolate a(1), subtract 4.44 from both sides:
-28 - 4.44 = a(1)
a(1) ≈ -32.44 (rounded to two decimal places).
Therefore, the rule for the nth term of the arithmetic sequence is:
a(n) = -32.44 + (n-1)(4.44)