**Question:**

Find the 12th term from the end of the following arithmetic progressions:

(i) 3, 5, 7, 9, … 201

(ii) 3, 8, 13, …, 253

(iii) 1, 4, 7, 10, …, 88

**Solution:**

In the given problem, we need to find the 12th term from the end for the given A.P.

(i) 3, 5, 7, 9 …201

Here, to find the 12th 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*) = 3

Last term (*a**n*) = 201

Common difference $(d)=5-3$

$=2$

Now, as we know,

$a_{s}=a+(n-1) d$

So, for the last term,

$201=3+(n-1) 2$

$201=3+2 n-2$

$201=1+2 n$

$201-1=2 n$

Further simplifying,

$200=2 n$

$n=\frac{200}{2}$

$n=100$

So, the 12th term from the end means the 89th term from the beginning.

So, for the 89th term (*n =* 89)

$a_{39}=3+(89-1) 2$

$=3+(88) 2$

$=3+176$

$=179$

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

(ii) 3, 8, 13 …253

Here, to find the 12th 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*) = 3

Last term (*a**n*) = 253

Common difference, $d=8-3$

$=5$

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$253=3+(n-1) 5$

$253=3+5 n-5$

$253=-2+5 n$

$253+2=5 n$

Further simplifying,

$255=5 n$

$n=\frac{255}{5}$

$n=51$

So, the 12th term from the end means the 40th term from the beginning.

So, for the 40th term (*n =* 40)

$a_{40}=3+(40-1) 5$

$=3+(39) 5$

$=3+195$

$=198$

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

(iii) 1, 4, 7, 10 …88

Here, to find the 12th 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*) = 1

Last term (*a**n*) = 88

Common difference, $d=4-1=3$

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$88=1+(n-1) 3$

$88=1+3 n-3$

$88=-2+3 n$

$88+2=3 n$

Further simplifying,

$90=3 n$

$n=\frac{90}{3}$

$n=30$

So, the 12th term from the end means the 19th term from the beginning.

So, for the 19th term (*n =* 19)

$a_{19}=1+(19-1) 3$

$=1+(18) 3$

$=1+54$

$=55$

Therefore, the $12^{\text {th }}$ term from the end of the given A.P. is 55 .