(i) $3,5,7,9,$ …. $201$

(ii) $3,8,13,$ … $,253$

An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

In order the find the ${{12}^{th}}$ term for the end of an A.P. which has n terms, its done by simply finding the $\left( \left( n-12 \right)+1 \right){}^{th}$ of the A.P

And we know, ${{n}^{th}}$ term ${{a}_{n}}=a+\left( n-1 \right)d$

(i) Given A.P $=3,5,7,9,$ …. $201$

Here, $a=3$ and $d=\left( 5-3 \right)=2$

Now, find the number of terms when the last term is known i.e, $201$

${{a}_{n}}=3+\left( n-1 \right)2=201$

$3+2-2=201$

$2n=200$

$n=100$

Hence, the A.P has $100$ terms.

So, the ${{12}^{th}}$ term from the end is same as ${{\left( 100-12+1 \right)}^{th}}$ of the A.P which is the ${{89}^{th}}$ term.

⇒ ${{a}_{89}}=3+\left( 89-1 \right)2$

$=3+88\left( 2 \right)$

$=3+176$

$=179$

Therefore, the ${{12}^{th}}$ term from the end of the A.P is $179.$

(ii) Given A.P $=3,8,13,$ … $,253$

Here, $a=3$ and $d=\left( 8-3 \right)=5$

Now, find the number of terms when the last term is known i.e, $253$

${{a}_{n}}=3+\left( n-1 \right)5=253$

$3+5n-5=253$

$5n=253+2=255$

$n=255/5$

$n=51$

Hence, the A.P has $51$ terms.

So, the ${{12}^{th}}$ term from the end is same as ${{\left( 51-12+1 \right)}^{th}}$ of the A.P which is the ${{40}^{th}}$ term.

⇒ ${{a}_{40}}=3+\left( 40-1 \right)15$

$=3+39\left( 5 \right)$

$=3+195$

$=198$

Therefore, the ${{12}^{th}}$ term from the end of the A.P is $198.$