how would one go about solving the Arithmetic Sequence of a1+a2+a3...+a98=2006?

Question

also, if a1+a2+a3+a4...+a98=2006, what will be the sum of a2+a3+a4+a5...+a98?

Answer 1

The 98 a values form an arithmetic sequence. That is, each successive one is larger than the previous one by a constant amount.

Let the first value be x. (That is, a1 = x.)
Let the increase amount be i.
So a2 = x + i, a3 = x + 2i, etc.

So the total for the 98 values is 98x + (i + 2i + 3i + ... + 97i).
(Note that a98 is the 97th term after a1, so it equals a1 plus 97i, not a1 plus 98i.)

We need a way to simplify the sum of the i's (from 1 i to 97 i's).
There is a formula for this. The sum is i (97)(98)/2 = 4753i.

So now we have:
98x + 4753i = 2006

There are lots of values of x and i that will satisfy this equation. In fact, you could choose ANY value for x and then solve for the value of i that will work: i = (2006 - 98x) / 4753

Are x and i supposed to be integers? If so, this problem is of a type called Diophantine equations. These are tricky to solve.
In this case, I believe that there is no integer solution, but I'm not positive.

Answer 2

Erm... do you mean a to the power of 1 and so on, or do you mean a, 2a, 3a?
If you are referring to a + 2a + 3a + ... + 98a = 2006,
pattern: a + 98a = 99a; 2a + 97a = 99a; 3a + 96a = 99a...
you will have 44 groups of 99a, ending with 44a + 45a = 99a
Thus 44(99a) = 2006
4356a = 2006
a = 0.461 (to 3 s.f.)

Hope this helps :)

Answer 3

add all the "a"s together and then divide 2006 by that number