# Find terms arithmetic progression

The nth term or general term of an A.P is used to find terms arithmetic progression.
Let us consider an A.P. with first term as 'a' and the common difference as 'd'.
So the sequence is a, a + d, a + 2d, a + 3d, ... then,
1st term = $a_{1}$ = a + (1 - 1)d
2nd term = $a_{2}$ = a + d = a + (2 -1)d
3rd term = $a_{3}$ = a + 2d = a + (3 -1)d
4th term = $a_{4}$ = a + 3d = a + (4 -1)d
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nth term = $a_{n}$ = a + (n- 1)d
From the above we observe that the terms of A.P is given by
First term + ( term number - 1 ) (common difference)
Formula to find nth term of an A.P sequence is

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

For example 15th term will be :
$a_{15}$ = a + (15 - 1)d = a + 14d

## Examples on Find terms arithmetic progression

1) Find the 19th and 24th terms of the A.P given by 21,16,11,6,1,...
Solution : We know that nth term of an A.P. is given by
$a_{n}$ = a + (n - 1) d
To find $a_{19}$,
n = 19, a = 21 and d = 16 - 21 = - 5
$a_{19}$ = 21 + ( 19 -1 )(-5)
= 21 + 18 x(-5)
= 21 - 90
∴ $a_{24}$ = - 69
To find $a_{24}$,
n = 24, a = 21 and d = 16 - 21 = - 5
$a_{24}$ = 21 + ( 24 -1 )(-5)
= 21 + 23 x(-5)
= 21 - 115
∴ $a_{24}$ = - 94

2) Which term of the A.P. is 1, 6,11,16,... is 301 ?
Solution : The given sequence is in A.P.
So, a = 1 and d = 6 -1 = 5
Let us consider nth term as 301
$a_{n}$ = a + (n - 1) d
301 = 1 + (n - 1)(5)
301 = 1 + 5n - 5
301 = -4 + 5n
301 + 4 = 5n
305 = 5n
∴ $\frac{305}{5} = \frac{5n}{5}$
∴ n = 61
Thus 301 is 61st term.

3) Which term of the A.P. 5,15,25,... will be 130 more than its 31st term.
Solution: We have, a = 5 and d = 15 - 5 = 10
$a_{n}$ = a + (n - 1)d
so, 31st term is
$a_{31}$ = 5 + (31 - 1)(10)
$a_{31}$ = 5 + 30 x 10
$a_{31}$ = 305
Let the nth term of the given A.P. be 130 more than its 31st term.
$a_{n}$ = 130 + $a_{31}$
$a_{n}$ = 130 + 305
$a_{n}$ = 435
5 + (n - 1)(10) = 435
5 + 10n - 10 = 435
- 5 + 10n = 430
10n = 440
$\frac{10n}{10} = \frac{440}{10}$
∴ n = 44
Hence, 44th term of the given A.P is 130 more than its 31st term.