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 $108$.