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

Question

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.

Answer 1

You haven't yet used the fact that the sum is $426$. Using the formula for the sum of an AP gives $$na+\frac{n(n-1)}{2}d=426.$$

Answer 2

Hint

$$\sum _{k=0}^n (a+5 k)=\frac{1}{2} (n+1) (2 a+5 n)=426$$ Should be $a=8$

Answer 3

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.