The $6^{t h}$ term from the end of the AP: 5, 2, -1, -4 ... ...., -31 is

-16

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-19

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-22

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-25

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Solution

The correct option is A -16
The given AP is 5, 2, -1, -4 ... ...., -31
d = 2 - 5 = -3, so d for the AP starting from the last term is 3.
The first term = l = -31
We know, $a_{n} = a + (n - 1) d$
$a_{6} f r o m t h e e n d = - 31 + (5) 3$
$a_{6} f r o m t h e e n d = - 16$

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