# Given the nth term for each arithmetic sequence, how to find the common difference and write out the first four terms here? (1) a_n=2n+7 (2) a_n=3n-2

Jun 27, 2018

See below.

#### Explanation:

Since every couple of consecutive terms in an arithmetic sequence differ by a common difference, we can subtract any two consecutive terms to find out how distant they are from each other.

So, in the first case, let's consider the ${n}^{\text{th}}$ and the $n + {1}^{\text{th}}$ term:

${a}_{n} = 2 n + 7 , \setminus q \quad {a}_{n + 1} = 2 \left(n + 1\right) + 7 = 2 n + 2 + 7 = 2 n + 9$

and subtract them:

${a}_{n + 1} - {a}_{n} = 2 n + 9 - \left(2 n + 7\right) = \cancel{2 n} + 9 - \cancel{2 n} - 7 = 2$

As for the first four terms, it depends if you consider the first term to be associated with $n = 0$ or $n = 1$. I usually go for the first choice. With this assumption, the first four terms are

${a}_{0} = 2 \cdot 0 + 7 = 0 + 7 = 7$
${a}_{1} = 2 \cdot 1 + 7 = 2 + 7 = 9$
${a}_{2} = 2 \cdot 2 + 7 = 4 + 7 = 11$
${a}_{3} = 2 \cdot 3 + 7 = 6 + 7 = 13$

In other words, you simply have to plug the value of $n$ in the expression for the general term.

The second sequence behaves exactly in the same way.