Find the 10th term from the end of the A.P. 8, 10, 12, ..., 126.

In the given problem, we need to find the 10th term from the end for the given A.P.

We have the A.P as 8, 10, 12 …126

Here, to find the 10th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 8

Last term (an) = 126

Common difference (d) = =2

Now, as we know,

So, for the last term,

$126=8+(n-1) 2$

$126=8+2 n-2$

$126=6+2 n$

$126-6=2 n$

Further simplifying,

$120=2 n$

$n=\frac{120}{2}$

$n=60$

So, the 10th term from the end means the 51st term from the beginning.

So, for the 51st term (n = 51)

$a_{54}=8+(51-1) 2$

$=8+(50) 2$

$=8+100$

$=108$

Therefore, the $10^{\text {th }}$ term from the end of the given A.P. is 108 .