Arithmetic Sequence with conditions

Question

The question is:

Let $\{a_n\}$ be an arithmetic sequence with the following conditions: $S_{16} = 376$ and $a_{16} = 46$, where $S_n$ is the sum of terms $a_1$ through $a_n$.

Find $a_1$.

Answer 1

Using the two relevant formula for APs you have:

$$376=S_n=16\cdot a_1+\frac{16\cdot15}{2}d=16\cdot a_1+120d$$

$$46=a_n=a_1+15d$$

The first can be simplified to: $$\frac{47}{2}=a_1+\frac{7}{2}d$$ The second one minus this gives: $$\frac{45}{2}=\frac{15}{2}d$$ $$d=3$$ Subbing $d$ back in gives: $$a=1$$

Answer 2

HINT:

Use the formula

$$S_n=\frac n2(a_1+a_n)$$

where $S_n$ is the sum of the first $n$ terms, $a_1$ is the first term, and $a_n$ is the $n$th term.

The given data gives the values of all variables in that equation except $a_1$, so solve for it.