# Given the first term and the common difference of an arithmetic sequence how do you find the first five terms and explicit formula: a1 = 3/5, d= -1/3?

Sep 19, 2015

${a}_{n} = {a}_{1} + \left(n - 1\right) d$
$A . P \left(\frac{9}{15} , \frac{4}{15} , - \frac{1}{15} , - \frac{6}{15} , - \frac{11}{15}\right)$

#### Explanation:

The concept of an arithmetic sequence is that, from the first term, every term after has a common difference between, so

${a}_{n} - {a}_{n - 1} = {a}_{n - 1} - {a}_{n - 2} = \ldots = {a}_{3} - {a}_{2} = {a}_{2} - {a}_{1} = d$

Which means, that we can rewrite ${a}_{2}$ as ${a}_{1} + d$, and we can also rewrite ${a}_{3}$ as ${a}_{2} + d$ or ${a}_{1} + 2 d$ and so on, until we start seeing a pattern. ${a}_{n} = {a}_{1} + \left(n - 1\right) d$, because we can put ${a}_{n - 1}$ in terms of ${a}_{n - 2}$ and that in terms of ${a}_{n - 3}$ and so on down to ${a}_{1}$, doing a total of $\left(n - 1\right)$ rewrites.

As for the terms, it's just a matter of evaluating in your matter of choice.
${a}_{1} = \frac{3}{5} = \frac{9}{15}$
${a}_{2} = \frac{3}{5} - \frac{1}{3} = \frac{9}{15} - \frac{5}{15} = \frac{4}{15}$
${a}_{3} = \frac{3}{5} - \frac{2}{3} = \frac{9}{15} - \frac{10}{15} = - \frac{1}{15}$
${a}_{4} = \frac{3}{5} - \frac{3}{3} = \frac{9}{15} - \frac{15}{15} = - \frac{6}{15}$
${a}_{5} = \frac{3}{5} - \frac{4}{3} = \frac{9}{15} - \frac{20}{15} = - \frac{11}{15}$