Question

Find the first term and the common difference of the arithmetic sequence described. Give a recursive formula for the sequence. Find a formula for the nth term.
$12$th term is $4$; $18$th term is$28$

Short answer

The solution is$a_1=-40$and$d=4$;$a_1=-40,a_n=a_{n-1}+4$;$a_n=4n-44$.

Step-by-step solution

Step 1. Given information 

$12$th term is $4$; $18$th term is$28$

Step 2. Calculation 

We know that $a_n=a_1+\left(n-1\right)d$. As a result,

$a_{12}=a_1+11d=4$ $\to 1$

$a_{18}=a_1+17d=28$ $\to 2$

Subtract the second equation from the first, we get -

$-6d=-24$

$d=4$

Substitute $d=4$in $a_1=4-11d=4-11·4=-40$.

Therefore the first term is $a_1=-40$and the common difference is $d=4$.

Step 3. Calculation 

Using formula -

$a_1=a,a_n=a_{n-1}+d$

a recursive formula for the sequence is -

$a_1=-40,a_n=a_{n-1}+4$

Step 4. Calculation 

Using formula -

$a_n=a_1+\left(n-1\right)d$

a formula for the $n$th term of the sequence is -

role="math" localid="1648731039528" $a_n=-40+\left(n-1\right)4$

$=-40+4n-4$

$=4n-44$