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Find the 10th term from the end of the A.P. $8, 10, 12, …, 126$.
Given:
Given A.P. is $8, 10, 12, …, 126$.
To do:
We have to find the 10th term from the end of the given arithmetic progression.
Solution:
In the given A.P.,
$a_1=8, a_2=10, a_3=12$
First term $a_1 = a= 8$, last term $l = 126$
Common difference $d = a_2-a_1 = 10 - 8 = 2$
We know that,
nth term from the end is given by $l - (n - 1 ) d$.
Therefore,
10th term from the end $= 126 - (10 - 1) \times 2 = 126 - 9 \times 2 = 126 - 18 = 108$.
The 10th term from the end of the given A.P. is $108$.   
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