Question

In Problems 20 and 21, find a general formula for each arithmetic sequence

7th term is 31;

20th term is 96

Short answer

The general formula of the sequence is :$a_n=5n-4$

Step-by-step solution

Step 1. Given information

In an arithmetic sequence :

$$\begin{gathered} a_7=31 \\ {a}_{20}=96 \end{gathered}$$

Let $a$is the first term and $d$is the common difference of the sequence.

Step 2. Make two equations of a and d

According to the given condition,

$$\begin{gathered} a_7=31 \\ a+(7-1)d=31 \end{gathered}$$

$a+6d=31$........... equation (1)

$$\begin{gathered} a_{20}=96 \\ a+(20-1)d=96 \end{gathered}$$.

$a+19d=96$...............equation (2)

Step 3. Solve equations.

Substract equation (1) from (2)

$$\begin{gathered} (a+19d)-(a+6d)=96-31 \\ 19d-6d=65 \\ 13d=65 \\ d=\frac{65}{13} \\ d=5 \end{gathered}$$

Substitute value of $d$in equation (1)

$$\begin{gathered} a+6\times 5=31 \\ a+30=31 \\ a=31-30 \\ a=1 \end{gathered}$$

Step 4. General formula for sequence.

Substitute values of $a$and $d$in general formula of arithmetic sequence.

$$\begin{gathered} a_n=a+(n-1)d \\ {a}_n=1+(n-1)5 \\ {a}_n=1+5n-5 \\ {a}_n=5n-4 \end{gathered}$$