Arithmetic Progression I

nth term of an A.P. is
tn = a + (n-1)d = 3n + 9
when n=1, t1 = 3 + 9 = 12 which is the first term
when n=2, t1 = 6 + 9 = 15
when n=2, t1 = 9 + 9 = 18
Therefore, common difference is 3

Given a = -6, d=-3
Arithmetic Progression is a, a + d, a + 2d, a + 3d,…
In this case, we get
-6, -6 + (-3), -6 + 2*(-3), -6 + 3*(-3),…..
-6, -9, -12, -15,…..

We have, first term a = 5, common difference d = 4
nth term of an A.P. is
tn = a + (n-1)d
16th term of an A.P. is
t16 = 5 + (16-1)4 = 5 + 60 = 65

a = 15, d = 3
Arithmetic Progression is a, a + d, a + 2d, a + 3d,…
In this case, we get
15, 15 + 3, 15 + 6,…
i.e. 15, 18, 21, 24, 27,…

Let the numbers be a-d, a, a+d
Given a = 11
(a-d)a(a+d) = 1155
a(a^2 – d^2) = 1155
121 – d^2 = 105
d = sqrt (16) = 4

Given a = 2
nth term of an A.P. is
tn = a + (n-1)d
a+(6-1)d = 4[a+(3-1)d] + 3
2 + 5d = 4 [2+2d] + 3
3d = -9
d = -3

Given, a=5, d=4, tn = 33
To find n
nth term of an A.P. is
tn = a + (n-1)d
33 = 5 + (n-1) 4
n = (33-5)/4 + 1 = 8

nth term of an A.P. is
tn = a + (n-1)d
6th term is 32
32 = a + (6-1)d = a+5d——(1)
12th term is 68
68 = a + (12-1)d = a+11d——(2)
(2) – (1) gives
36 = 6d
d = 6

The formula for the nth term of an AP is:
an=a+(n−1)⋅da_n = a + (n – 1) \cdot dan​=a+(n−1)⋅d
Where:

• ana_nan​ is the nth term,
• $a$ is the first term,
• $d$ is the common difference, and
• $n$ is the term number.

Substitute the given values:
$a=3$, $d=5$, $n=10$.

a10=3+(10−1)⋅5=3+9⋅5=3+45=48a_{10} = 3 + (10 – 1) \cdot 5 = 3 + 9 \cdot 5 = 3 + 45 = 48a10​=3+(10−1)⋅5=3+9⋅5=3+45=48