Question

I need help with this question its on arithmetic growthanddecay , been stuck on it for many days and i need help! pleasep

Answer 1

We are given the following information about the arithmetic sequence

First two rows = 27 chairs

Last two rows = 114 chairs

Common difference = 3 chairs

Recall that the general formula for an arithmetic sequence is given by

`a_n=a_1+(n-1)d`

(a) Let us substitute the given values into the above formula and solve for n

`\begin{gathered} a_n=a_1+(n-1)d \\ 114=27_{}+(n-1)\cdot3 \\ 114-27=_{}(n-1)\cdot3 \\ 87=_{}(n-1)\cdot3 \\ (87)/(3)=_{}n-1 \\ 29=_{}n-1 \\ 29+1_{}=_{}n \\ 30=n \end{gathered}`

There are 30 rows of chairs.

(b) Let us find the number of chairs in the 13th and 30th row.

i) 13th row:

Substitute n = 13

`\begin{gathered} a_(13)=27_{}+(13-1)\cdot3 \\ a_(13)=27_{}+12\cdot3 \\ a_(13)=27_{}+36 \\ a_(13)=63 \end{gathered}`

There are 63 chairs in the 13th row.

ii) 30th row:

Substitute n = 30

`\begin{gathered} a_(30)=27_{}+(30-1)\cdot3 \\ a_(30)=27_{}+29\cdot3 \\ a_(30)=27_{}+87 \\ a_(30)=114 \end{gathered}`

There are 114 chairs in the 30th row.