Find the $15 t h$ term of an AP. $y - 7, y - 2, y + 3, . . . . . . . .$ .

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Solution

$a = y - 7$
$d = a_{2} - a_{1}$
$= y - 2 - (y - 7)$
$= y - 2 - y + 7$
$= 5$
$a_{n} = a + (n - 1) d$
$a_{15} = (y - 7) + (15 - 1) 5$
$a_{15} = (y - 7) + 14.5$
$a_{15} = y - 7 + 70$
$a_{15} = y + 63$ .

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