Question

What is the 32nd term of the arithmetic sequence where a1 = 14 and a13 = −58? (1 point)−178−172−166−160

Answer 1

Step 1

Given;

$\begin{gathered} First\text{ term\lparen a}_1)=14 \\ a_{13}=-58 \end{gathered}$

Step 2

Find the 32nd term

State the nth term of an A.P

$\begin{gathered} a_n=a_1+(n-1)d \\ \end{gathered}$

Find the equation for the 13th term

$\begin{gathered} a_{13}=14+(13-1)d \\ -58=14+12d \end{gathered}$

Find d, the common difference

$\begin{gathered} -58=14+12d \\ 12d=-58-14 \\ \frac{12d}{12}=-\frac{72}{12} \\ d=-6 \end{gathered}$

Step 3

Find the 32nd term

$\begin{gathered} a_{32}=a_1+(32-1)d \\ a_{32}=14+31(-6) \\ a_{32}=-172 \end{gathered}$

Answer;

$The\text{ 32nd term=-172}$