# How do I find the first term in an arithmetic sequence?

The last term of an arithmetic sequence is $$63$$, the common difference is $$5$$ and the sum is $$426$$. How do you find the first term?

That's all the information I have.

My attempt:

$$63 = a + (n-1)(5)$$

$$63 = a + 5n - 5$$

$$68 = a + 5n$$

I'm stuck here.

You haven't yet used the fact that the sum is $$426$$. Using the formula for the sum of an AP gives $$na+\frac{n(n-1)}{2}d=426.$$
$$\sum _{k=0}^n (a+5 k)=\frac{1}{2} (n+1) (2 a+5 n)=426$$ Should be $$a=8$$
Since you were given the last term, you can also apply the sum formula $$\ s \ = \ \frac{n}{2}·(a_1 + a_n) \ \ ,$$ which produces $$\ 426 \ = \ \frac{n}{2}·(a_1 + 63) \ \Rightarrow \ ( a_1 + 63 ) · n \ = \ 852 \ \ .$$ Putting this together with the equation you found, you can substitute either $$a_1 \ \ = \ \ 68 \ - \ 5n \ \ \Rightarrow \ \ ( 131 \ - \ 5n ) · n \ = \ 852 \ \ \Rightarrow \ \ 5n^2 \ - \ 131n \ + \ 852 \ \ = \ \ 0$$ or $$n \ \ = \ \ \frac{68 \ - \ a_1}{5} \ \ \Rightarrow \ \ ( 63 \ + \ a_1 ) · \left( \frac{68 \ - \ a_1}{5} \right) \ = \ 852 \ \ \Rightarrow \ \ a_1^2 \ - \ 5a_1 \ - \ 24 \ \ = \ \ 0 \ \ .$$
Either choice leads to a quadratic equation. The latter is easier to factor, yielding $$\ (a_1 - 8)·(a_1 + 3) \ = \ 0 \ \ ,$$ from which we obtain $$\mathbf{a_1 \ = \ 8 \ \ : \ } \ \ n \ \ = \ \ \frac{68 \ - \ 8}{5} \ \ = \ \ 12 \ \ \ \text{or} \ \ \ \mathbf{a_1 \ = \ -3 \ \ : \ } \ \ n \ \ = \ \ \frac{68 \ - \ [-3]}{5} \ \ = \ \ 14.2 \ \ ,$$ so only the first result is admissible. Checking the sum by the usual formula, we obtain $$\ 12·8 \ + \ \frac{12·11}{2}·5 \ = \ 96 + 330 \ = \ 426 \ \ .$$ The first quadratic equation has a discriminant of $$\ 121 \ \ ,$$ so we have $$\ n \ = \ \frac{131 \ \pm \ 11}{2·5} \ = \ \frac{120}{10} \ = \ 12 \ \ , \ \ \frac{142}{10} \ = \ 14.2 \ \ ;$$ the integer number of terms tells us that $$\ a_1 \ \ = \ \ 68 \ - \ 5·12 \ = \ 8 \ \ ,$$ which we have found to produce the correct sum.