Find $20^{t h}$ term from the end of an AP 3, 7, 11, ......, 407

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Solution

$a_{n} = a + (n - 1) d$
Here $a_{n} = 407, a = 3, d = 4$
$407 = 3 + (n - 1) 4$
$407 = 3 + 4 n - 4$
$408 = 4 n$
$\Rightarrow n = 102$
$∴ 20^{t h}$ term from end $\Rightarrow m = 20$
$a_{102 - (20 - 1)} = a_{102 - 19} = a_{83}$ from the beginning
$a_{83} = 3 + (83 - 1) 4 = 331$

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