arithmetic sequence Question?

The general for for the n'th term of an arithmetic sequence is:

a(n) = a + b(n - 1)

Where a is the first term and b is the common difference.

Then we have an equation that can give us the sum of the first "n" terms of a sequence:

S(n) = [ a + a(n)] n / 2

We are told that the sum of the first 5 terms is 85 and the sum of the first 6 terms is 123, so let's find expressions for a(5) and a(6), first:

a(n) = a + b(n - 1)
a(5) = a + b(5 - 1) and a(6) = a + b(6 - 1)
a(5) = a + 4b and a(6) = a + 5b

We don't yet know what a(5) and a(6) are, so leave them at that.

Now we can find expressions for S(5) and S(6):

S(n) = [ a + a(n)] n / 2
S(5) = [ a + a(5)] * 5 / 2 and S(6) = [ a + a(6)] * 6 / 2
S(5) = 5[ a + a(5)] / 2 and S(6) = 3[ a + a(6)]

We do have expressions for a(5) and a(6), so:

S(5) = 5(a + a + 4b) / 2 and S(6) = 3(a + a + 5b)
S(5) = 5(2a + 4b) / 2 and S(6) = 3(2a + 5b)
S(5) = (10a + 20b) / 2 and S(6) = 6a + 15b
S(5) = 5a + 10b and S(6) = 6a + 15b

And we do know the values for S(5) and S(6):

85 = 5a + 10b and 123 = 6a + 15b

Now we have a system of two equations and two unknowns so we can solve for "a" and "b" (the first term and the common difference). Once we know those, we can find the values of the first 4 terms.

Solving the first equation for a in terms of b:

85 = 5a + 10b
85 - 10b = 5a
17 - 2b = a

Substitute into the other equation to solve for b:

123 = 6a + 15b
123 = 6(17 - 2b) + 15b
123 = 102 - 12b + 15b
123 = 102 + 3b
21 = 3b
b = 7

Now we can solve for a:

a = 17 - 2b
a = 17 - 2(7)
a = 17 - 14
a = 3

So the first four terms of this sequence is:

3, 10, 17, 24

a + 5d = 123
a + 4d = 86
d = 37
a = 86 – 4*37 = -62

The first 4 terms are:-
-62, -25, 12, 49