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 $-78$.
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