Determine the A.P whose fourth term is $18$ and the difference of the ninth term from the fifteenth term is $30$ .

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Solution

It is given that 4th term is 18
$a_{4} = 18$
$a + 3 d = 18$
$a = 18 - 3 d . . . . . . . (1)$
Also
$a_{15} - a_{9} = a + 14 d - a - 8 d = 30$
$6 d = 30$
$d = 5$
Thus, form (1), we get
$a = 18 - 15 = 3$
Therefore, the AP is 3,8,13,18,23....

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