# How do you write the nth term rule for the arithmetic sequence with a_5=17 and a_15=77?

Jul 28, 2016

${a}_{n} = 6 n - 13 , n \in \mathbb{N}$.

#### Explanation:

Let ${a}_{1} , {a}_{n} \mathmr{and} d \ne 0$ denote the first , ${n}^{t h}$ and common difference for the Arithmetic Sequence in question.

Then, we know that, ${a}_{n} = {a}_{1} + \left(n - 1\right) d \ldots \ldots \ldots . . \left(1\right)$

Given that ${a}_{5} = 17$, so, taking $n = 5$ in $\left(1\right)$, we have,

${a}_{1} + 4 d = 17. \ldots \ldots \ldots \ldots . . \left(2\right)$.

Similarly, ${a}_{15} = 77 \Rightarrow {a}_{1} + 14 d = 77. \ldots \ldots \ldots \ldots \ldots . \left(3\right)$

Solving (2) & (3), d=6, and, a_1=-7.

These, with $\left(1\right)$ gives ${a}_{n} = 6 n - 13 , n \in \mathbb{N}$.