Question

Given two terms of the arithmetic sequence, a_6=32 and a_10= -8, find a_1 and d.

Answer 1

`\begin{gathered} a_1=82 \\ d=-10 \end{gathered}`

Step-by-step explanation

In an Arithmetic Sequence the difference between one term and the next is a constant. In other words, we just add the same value each time , it is given by the expression

`\begin{gathered} a_n=a_(1+)(n-1)d \\ \text{where} \\ a_1i\text{s the first term } \\ and\text{ } \\ d\text{ is the common differnece, so} \end{gathered}`

Step 1

Set the equations:

a)

Let

`\begin{gathered} a_6=a_1+(6-1)d \\ \text{replace} \\ 32=a_1+5d\rightarrow equation(1) \end{gathered}`

b)

`\begin{gathered} a_{_(10)}=a_1+(10-1)d \\ \text{replace} \\ -8=a_1+9d\rightarrow equation(2) \end{gathered}`

Step 2

Now, solve the equations

`\begin{gathered} 32=a_1+5d\rightarrow equation(1) \\ -8=a_1+9d\rightarrow equation(2) \end{gathered}`

a) isolate a1 in equation (1) and then replace in equation (2)

`\begin{gathered} 32=a_1+5d\rightarrow equation(1) \\ \text{subtract 5d in both sides} \\ 32-5d=a_1+5d-5d \\ 32-5d=a_1 \end{gathered}`

replace i equation (2)

`\begin{gathered} -8=a_1+9d\rightarrow equation(2) \\ -8=32-5d+9d \\ \text{add like terms} \\ -8=32+4d \\ \text{subtract 32 in both sides} \\ -8-32=32+4d-32 \\ -40=4d \\ \text{divide both sides by 4} \\ (-40)/(4)=(4d)/(4) \\ -10=d \\ d=-10 \end{gathered}`

b) now , fo find a1, replace d in equation(1)

`\begin{gathered} 32=a_1+5d\rightarrow equation(1) \\ 32=a_1+5(-10) \\ 32=a_1-50 \\ \text{add 50 in both sides} \\ 32+50=a_1-50+50 \\ 82=a_1 \end{gathered}`

therefore, the answer is

`\begin{gathered} a_1=82 \\ d=-10 \end{gathered}`

I hope this helps you