Find the $ 12^{\text {th }} $ term from the end of the AP: $ -2,-4,-6, \ldots,-100 $.

Given:

Given A.P. is $-2, -4, -6, …, -100$.

To do:

We have to find the 12th term from the end of the given arithmetic progression. 

Solution:

In the given A.P.,

$a_1=-2, a_2=-4, a_3=-6$

First term $a_1 = a= -2$, last term $l = -100$

Common difference $d = a_2-a_1 = -4 - (-2) = -4+2=-2$

We know that,

nth term from the end is given by $l - (n - 1 ) d$.

Therefore,

12th term from the end $= -100 - (12 - 1) \times (-2)$

$= -100 - 11 \times (-2)$

$= -100 + 22$

$= -78$.

The 12th term from the end of the given A.P. is $-78$.