Answer
The first ten terms of the arithmetic sequence is 50, 58, 66, 74, 82, 90, 98, 106, 114 and 122
The number of seats in the tenth row is 122.
Work Step by Step
For the arithmetic sequence, $a_{1} = 50$ and $d = 8$, the general term $a_{n} = 50 + (n - 1)(8)$ where n is the number of rows.
The first ten terms of the sequence are respectively
$a_{1} = 50 + (1 - 1)(8) = 50$
$a_{2} = 50 + (2 - 1)(8) = 58$
$a_{3} = 50 + (3 - 1)(8) = 66$
$a_{4} = 50 + (4 - 1)(8) = 74$
$a_{5} = 50 + (5 - 1)(8) = 82$
$a_{6} = 50 + (6 - 1)(8) = 90$
$a_{7} = 50 + (7 - 1)(8) = 98$
$a_{8} = 50 + (8 - 1)(8) = 106$
$a_{9} = 50 + (9 - 1)(8) = 114$
$a_{10} = 50 + (10 - 1)(8) = 122$
The first ten terms of the arithmetic sequence is 50, 58, 66, 74, 82, 90, 98, 106, 114 and 122
And the number of seats in the tenth row is 122.
