Let  $a_1,a_2,a_3, ....$ be  an  AP  whose  first  term  $a_1$ is $a$ and the common difference $d$.


Then, 
the  second  term  $a_2=  a + d = a + (2 – 1) d $
the third term   $a_3=  a_2+ d = (a + d) + d = a + 2d = a + (3 – 1) d$
the fourth term  $a_4=a_3+d= (a + 2d) + d = a + 3d = a + (4 – 1) d $
So, $a_n$ term  $a_n=a_{n-1}+d=a+(n-1)d $

Example: Find the 10th term of the AP :  2, 7, 12, . . .
Solution : Here, a = 2,    d = 7 – 2 = 5    and    n =  10. 
We have            $a_n  = a + (n – 1) d.$ 
So,               $a_10$ = 2 + (10 – 1) × 5 = 2 + 45 = 47 

Therefore, the 10th term of the given AP is 47.



Example: Which term of the AP : 21, 18, 15, . . . is  – 81? Also, is any term 0? Give reason for your answer.
Solution : Here, $a = 21,  d = 18 – 21 = – 3 $ and  $a_n  = – 81$, and we have to find $n$. As                    
             $a_n  = a + (n – 1) d.$ ,
we have       – 81 =  21 + (n – 1)(– 3)
              – 81 =  24 – 3n
			  – 105 =  – 3n

So,              n =  35
Therefore, the 35th term of the given AP is – 81.

Next, we want to know if there is any $n$ for which $a_n  = 0$. If such an $n$ is there, then
21 + (n – 1) (–3) =  0, 
i.e., 3(n – 1) =  21 
i.e.,   n =  8
So, the eighth term is 0.

Example: Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution : Since $a_n$ term is given by the equation,
         $a_n  = a + (n – 1) d.$
         $a_3= a + (3 – 1) d =  a + 2d = 5  $                              (1)
         $a_7= a + (7 – 1) d =  a + 6d = 9 $                                    (2) 
Solving the pair of linear equations (1) and (2), we get
$a =  3,    d = 1$
Hence, the required AP is 3, 4, 5, 6, 7, . . . 

Example: Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .
Solution : We have : 
$a_2-a_1$=  11 – 5 = 6, $a_3-a_2=17-11=6$,$a_4-a_3=23-17=6$
  
As $a_{k+1}-a_k$ is the same for $k = 1, 2, 3$, etc., the given list of numbers is an AP.
 Now,                  $a = 5 $   and        $d = 6$.
Let 301 be a term, say, the nth term of this AP.
We know that
          $a_n  = a + (n – 1) d.$
So,      301 =  5 + (n – 1) × 6
i.e.,    301 =  6n – 1
 
So,       $n = 302 /6$

But $n$ should be a positive integer (Why?). So, 301 is not a term of the given list of numbers.

Example: How many two-digit numbers are divisible by 3?
Solution : The list of two-digit numbers divisible by 3 is :
	12, 15, 18, . . . , 99
	Since the series is in AP, Hence,    $a = 12,  d = 3,  a_n  = 99$. 
	As $a_n  = a + (n – 1) d.$
       $99 =  12 + (n – 1) × 3$
	   i.e.,87 =  (n – 1) × 3

i.e.,   $n – 1 = 87/3 = 29$
i.e.,   n =  29 + 1 = 30
So, there are 30 two-digit numbers divisible by 3.


Example: Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62.

Solution : Here,    $a = 10,  d = 7 – 10 = – 3,  l = – 62$, where                                                  
             $l =  a + (n – 1) d$ 
To find the 11th term from the last term, we will find the total number of terms in the AP.
So,       – 62 = 10 + (n – 1)(–3)
i.e.,     – 72 = (n – 1)(–3)
i.e.,    n – 1 = 24 or  n = 25
So, there are 25 terms in the given AP.
The 11th term from the last term will be the 15th term. (Note that it will not be the 14th term. Why?) 
(Hint: reverse count the terms from the 25th term i.e, 25th,24th,23th,22th,21st,20th,19th,18th,17th,16th,15th)
(In general, the $lth$ term from the last $nth$ term is given by the formula =$(n-l)+1$ )
So,  $a_15$ =  10 + (15 – 1)(–3) = 10 – 42 = – 32 
i.e., the 11th term from the last term is – 32.

Alternative  Solution  :
If we write the given AP in the reverse order, then a = – 62 and d = 3 (Why?) So, the question now becomes 
finding the 11th term with these $a$ and $d$. 
So,   $a$ =  – 62 + (11 – 1) × 3 = – 62 + 30 = – 32 
So, the 11th term, which is now the required term, is – 32.

Homework:A sum of ₹ 1000 is invested at 8% simple interest per year. Calculate the interest 
at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making 
use of this fact.
(Hint: Simple Interest=PxRxT/100)

Homework: In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, 
and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?