Write down the formula of two number of ap and gp
Answers
Arithmetic Progression(AP)
Arithmetic progression(AP) or arithmetic sequence is a sequence of numbers in which each term after the first is obtained by adding a constant, d to the preceding term. The constant d is called common difference.
An arithmetic progression is given by a, (a + d), (a + 2d), (a + 3d), ...
where a = the first term , d = the common difference
Examples
1, 3, 5, 7, ... is an arithmetic progression (AP) with a = 1 and d = 2
7, 13, 19, 25, ... is an arithmetic progression (AP) with a = 7 and d= 6
If a, b, c are in AP, 2b = a + c
nth term of an arithmetic progression
tn = a + (n – 1)d
where tn = nth term, a= the first term , d= common difference
Example 1
Find 10th term in the series 1, 3, 5, 7, ...
a = 1
d = 3 – 1 = 2
10th term, t10 = a + (n-1)d = 1 + (10 – 1)2 = 1 + 18 = 19
Example 2
Find 16th term in the series 7, 13, 19, 25, ...
a = 7
d = 13 – 7 = 6
16th term, t16 = a + (n-1)d = 7 + (16 – 1)6 = 7 + 90 = 97
Number of terms of an arithmetic progression
n
=
(
l
−
a
)
d
+
1
where n = number of terms, a= the first term , l = last term, d= common difference
Example
Find the number of terms in the series 8, 12, 16, . . .72
a = 8
l = 72
d = 12 – 8 = 4
n
=
(
l
−
a
)
d
+
1
=
(
72
−
8
)
4
+
1
=
64
4
+
1
=
16
+
1
=
17
Sum of first n terms in an arithmetic progression
S
n
=
n
2
[
2
a
+
(
n
−
1
)
d
]
=
n
2
(
a
+
l
)
where a = the first term,
d= common difference,
l
= tn = nth term = a + (n-1)d
Example 1
Find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms
a = 4
d = 7 – 4 = 3
Sum of first 20 terms, S20
=
n
2
[
2
a
+
(
n
−
1
)
d
]
=
20
2
[
(
2
×
4
)
+
(
20
−
1
)
3
]
=
10
[
8
+
(
19
×
3
)
]
=
10
(
8
+
57
)
=
650
Example 2
Find 6 + 9 + 12 + . . . + 30
a = 6
l = 30
d = 9 – 6 = 3
n
=
(
l
−
a
)
d
+
1
=
(
30
−
6
)
3
+
1
=
24
3
+
1
=
8
+
1
=
9
Sum, S
=
n
2
(
a
+
l
)
=
9
2
(
6
+
30
)
=
9
2
×
36
=
9
×
18
=
162
Arithmetic Mean
If a, b, c are in AP, b is the Arithmetic Mean (AM) between a and c. In this case,
b
=
1
2
(
a
+
c
)
The Arithmetic Mean (AM) between two numbers a and b =
1
2
(
a
+
b
)
If a, a1, a2 ... an, b are in AP we can say that a1, a2 ... an are the n Arithmetic Means between a and b.
Additional Notes on AP
To solve most of the problems related to AP, the terms can be conveniently taken as
3 terms: (a – d), a, (a +d)
4 terms: (a – 3d), (a – d), (a + d), (a +3d)
5 terms: (a – 2d), (a – d), a, (a + d), (a +2d)
Tn = Sn - Sn-1
If each term of an AP is increased, decreased , multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.
In an AP, sum of terms equidistant from beginning and end will be constant.
Harmonic Progression(HP)
Non-zero numbers
a
1
,
a
2
,
a
3
,
⋯
a
n
are in Harmonic Progression(HP) if
1
a
1
,
1
a
2
,
1
a
3
,
⋯
1
a
n
are in AP. Harmonic Progression is also known as harmonic sequence.
Examples
1
2
,
1
6
,
1
10
,
⋯
is a harmonic progression (HP)
Three non-zero numbers a, b, c will be in HP, if
1
a
,
1
b
,
1
c
are in AP
If a, (a+d), (a+2d), . . . are in AP, nthterm of the AP = a + (n - 1)d
Hence, if
1
a
,
1
a
+
d
,
1
a
+
2
d
,
⋯
are in HP, nthterm of the HP =
1
a
+
(
n
−
1
)
d
If a, b, c are in HP, b is the Harmonic Mean(HM) between a and c
In this case,
b
=
2
a
c
a
+
c
The Harmonic Mean(HM) between two numbers a and b =
2
a
b
a
+
b
If a, a1, a2 ... an, b are in HP we can say that a1, a2 ... an are the n Harmonic Means between a and b.
If a, b, c are in HP,
2
b
=
1
a
+
1
c
geometric progression(GP)
Geometric Progression(GP) or Geometric Sequence is sequence of non-zero numbers in which the ratio of any term and its preceding term is always constant.
A geometric progression(GP) is given by a, ar, ar2, ar3, ...
where a = the first term , r = the common ratio
Examples
1, 3, 9, 27, ... is a geometric progression(GP) with a = 1 and r = 3
2, 4, 8, 16, ... is a geometric progression(GP) with a = 2 and r = 2
If a, b, c are in GP, b2 = ac
nth term of a geometric progression(GP)
t
n
=
a
r
n
−
1
where tn = nth term, a= the first term , r = common ratio, n = number of terms
Example 1
Find the 10th term in the series 2, 4, 8, 16, ...
a = 2, r =
4
2
= 2, n = 10
10th term, t10
=
a
r
n
−
1
=
2
×
2
10
−
1
=
2
×
2
9
=
2
×
512
=
1024
Example 2
Find 5th term in the series 5, 15, 45, ...
a = 5, r =
15
5
= 3, n = 5
5th term, t5
=
a
r
n
−
1
=
5
×
3
5
−
1
=
5
×
3
4
=
5
×
81
=
405
Sum of first n terms in a geometric progression(GP)
S
n
=
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
a
(
r
n
−
1
)
r
−
1
(
if
r
>
1
)
a
(
1
−
r
n
)
1
−
r
(
if
r
<
1
)
where a= the first term,
r = common ratio,
n = number of terms
Example 1
Find 4 + 12 + 36 + ... up to 6 terms
a = 4, r =
12
4
= 3, n = 6
Here r > 1. Hence,
S
6
=
a
(
r
n
−
1
)
r
−
1
=
4
(
3
6
−
1
)
3
−
1
=
4
(
729
−
1
)
2
=
4
×
728
2
=
2
×
728
=
1456
Example 2
Find
1
+
1
2
+
1
4
+
... up to 5 terms
a = 1, r =
(
1
2
)
1
=
1
2
,
n = 5
Here r < 1. Hence,
S
6
=
a
(
1
−
r
n
)
1
−
r
=
1
[
1
−
(
1
2
)
5
]
(
1
−
1
2
)
=
(
1
−
1
32
)
(
1
2
)
=
(
31
32
)
(
1
2
)
=
31
16
=
1
15
16
Sum of an infinite geometric progression(GP)
S
∞
=
a
1
−
r
(if -1 < r < 1)
where a= the first term , r = common ratio
Example
Find
1
+
1
2
+
1
4
+
1
8
+
⋯
∞
a = 1, r =
(
1
2
)
1
=
1
2
Here -1 < r < 1. Hence,
S
∞
=
a
1
−
r
=
1
(
1
−
1
2
)
=
1
(
1
2
)
=
2