Find the 12th term from the end of the following arithmetic progressions:$1, 4, 7, 10, …, 88$

Given:

Given A.P. is $1, 4, 7, 10,…, 88$.
To do:

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

In the given A.P.,

$a_1=1, a_2=4, a_3=7$

First term $a_1 = a= 1$, last term $l = 88$

Common difference $d = a_2-a_1 = 4 - 1 = 3$

We know that,

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

Therefore,

12th term from the end $= 88 - (12 - 1) \times 3 = 88 - 11 \times 3 = 88 - 33 = 55$.

The 12th term from the end of the given A.P. is $55$.