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 \\ (12d)/(12)=-(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}`