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
[tex]a_n=a_1+(n-1)d[/tex]
(a) Let us substitute the given values into the above formula and solve for n
[tex]\begin{gathered} a_n=a_1+(n-1)d \\ 114=27_{}+(n-1)\cdot3 \\ 114-27=_{}(n-1)\cdot3 \\ 87=_{}(n-1)\cdot3 \\ \frac{87}{3}=_{}n-1 \\ 29=_{}n-1 \\ 29+1_{}=_{}n \\ 30=n \end{gathered}[/tex]
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
[tex]\begin{gathered} a_{13}=27_{}+(13-1)\cdot3 \\ a_{13}=27_{}+12\cdot3 \\ a_{13}=27_{}+36 \\ a_{13}=63 \end{gathered}[/tex]
There are 63 chairs in the 13th row.
ii) 30th row:
Substitute n = 30
[tex]\begin{gathered} a_{30}=27_{}+(30-1)\cdot3 \\ a_{30}=27_{}+29\cdot3 \\ a_{30}=27_{}+87 \\ a_{30}=114 \end{gathered}[/tex]
There are 114 chairs in the 30th row.
