In AP $a_{3} = 21, a, 3 = 21, a_{1} 0 = 63$ find $a_{50}$

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Solution

$a_{1} = a = 63, a_{3} \neq a + 2 d = 21 \Rightarrow 2 d = 21 - 63$
$\Rightarrow d = - 42 / 2 = - 21$
$\Rightarrow a_{50} = a + 49 d = 63 - (49 \times 21) = 21 (3 - 49) = 21 (- 46)$
$a_{50} = - 966$

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