For the answer to the question above,
So I'll do the computations and try my best to explain, but you're going to have to follow along.
For the sake of simplicity, lets just say that the 3rd term they gave you was >3, and the 25th is >25. So, instead of saying A sub-3 is 3 and A sub-25 is 25, you set 3 as A sub-one. Now, heres the tricky part. Set 25 to A sub-23. (why? because 1 to 23 is the same amount away as 3 to 25).
A3=3, A25=25 }
A1=3, A23=25
Use the formula An = A1 + (n-1)D.
(read A-sub n is A sub 1 + the nth term minus 1 times D)
A23(A sub-23)= 3 + (23-1)D
25 = 3 + 22D
-3 -3
22 = 22D
D = 1
Again, An = A1 + (n-1)D.
A3 = A1 + (3-1) X 1
3 = A1 + 2
-2 -2
A1 = 1
An = 1 + (n-1) X 1
An = 1 + 1n - 1
An = 1n + 0
So you are left with the general formula An = 1n. You can use this formula to find any term in the sequence, all you gotta do is plug in (An) the number you want. In this case, plug in A sub-5 if you want to find the fifth term. You get
An = 1n
A5 = 1X5
A5 = 5
The fifth term in the sequence is 5. (it goes 1,2,3,4,,5,..)
