Important Concepts and Formulas - Sequence and SeriesArithmetic 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

$n = \dfrac{(l - a)}{d} + 1$

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

$n=\dfrac{(l-a)}{d}+1=\dfrac{(72-8)}{4}+1\\=\dfrac{64}{4}+1=16+1=17$

Sum of first n terms in an arithmetic progression

$S_n = \dfrac{n}{2} [\ 2a + (n - 1)d\ ] \ = \dfrac{n}{2}(a+l)$

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
$= \dfrac{n}{2} [2a+(n-1)d]\\= \dfrac{20}{2} [(2 \times 4)+(20-1)3]\\= 10 [8 + (19 \times 3)]\\=10(8+57)\\=650$

Example 2
Find 6 + 9 + 12 + . . . + 30

a = 6
l = 30
d = 9 – 6 = 3

$n = \dfrac{(l - a)}{d} + 1 \\= \dfrac{(30 - 6)}{3} + 1 \\= \dfrac{24}{3} + 1 \\= 8 + 1 \\= 9$

Sum, S
$=\dfrac{n}{2}(a+l)\\= \dfrac{9}{2}(6+30)\\=\dfrac{9}{2} \times 36 \\= 9 \times 18 \\= 162$

Arithmetic Mean

If a, b, c are in AP, b is the Arithmetic Mean (AM) between a and c. In this case, $b = \dfrac{1}{2}(a + c)$

The Arithmetic Mean (AM) between two numbers a and b = $\dfrac{1}{2}(a + b)$
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.

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 $a_1,\ a_2,\ a_3,\ \cdots \ a_n$ are in Harmonic Progression(HP) if $\dfrac{1}{a_1},\ \dfrac{1}{a_2},\ \dfrac{1}{a_3},\ \cdots \dfrac{1}{a_n}$ are in AP. Harmonic Progression is also known as harmonic sequence.

Examples

$\dfrac{1}{2}, \dfrac{1}{6}, \dfrac{1}{10}, \cdots$ is a harmonic progression (HP)

Three non-zero numbers a, b, c will be in HP, if  $\dfrac{1}{a},\ \dfrac{1}{b},\ \dfrac{1}{c}$ are in AP
If a, (a+d), (a+2d), . . . are in AP, nthterm of the AP = a + (n - 1)d

Hence, if $\dfrac{1}{a}, \dfrac{1}{a+d}, \dfrac{1}{a+2d}, \cdots$ are in HP, nthterm of the HP = $\dfrac{1}{a + (n - 1)d}$
If a, b, c are in HP, b is the Harmonic Mean(HM) between a and c

In this case, $b = \dfrac{2ac}{a + c}$
The Harmonic Mean(HM) between two numbers a and b = $\dfrac{2ab}{a + b}$
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, $\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
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)

$t_n = ar^{n-1}$

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 = $\dfrac{4}{2}$ = 2,   n = 10

10th term, t10
$=ar^{n-1} = 2 \times 2^{10-1} \\= 2 \times 2^{9} = 2 \times 512 = 1024$

Example 2
Find 5th term in the series 5, 15, 45, ...

a = 5,   r = $\dfrac{15}{5}$ = 3,   n = 5

5th term, t5
$=ar^{n-1} = 5 \times 3^{5-1}\\= 5 × 3^4 = 5 × 81 = 405$

$S_n =\begin{cases} \dfrac{a(r^n - 1)}{r - 1}\ & (\text{if } r \gt 1) \\ \\ \dfrac{a(1 - r^n)}{1 - r}\ & (\text{if } r \lt 1) \\\end{cases}$

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 = $\dfrac{12}{4}$ = 3,   n = 6

Here r > 1. Hence,
$S_6= \dfrac{a(r^n - 1)}{r - 1} = \dfrac{4(3^6 - 1)}{3 - 1} \\= \dfrac{4(729 - 1)}{2}= \dfrac{4 \times 728}{2} \\= 2 \times 728 = 1456$

Example 2
Find $1 + \dfrac{1}{2} + \dfrac{1}{4} +$ ... up to 5 terms

a = 1,   r = $\dfrac{\left(\dfrac{1}{2}\right)}{1} = \dfrac{1}{2},$   n = 5

Here r < 1. Hence,
$S_6 = \dfrac{a(1 - r^n)}{1 - r} = \dfrac{1\left[1 - \left(\dfrac{1}{2}\right)^5 \right]}{\left(1 - \dfrac{1}{2}\right)} \\= \dfrac{\left(1 - \dfrac{1}{32} \right)}{\left(\dfrac{1}{2}\right)} = \dfrac{\left(\dfrac{31}{32}\right) }{\left(\dfrac{1}{2}\right)} = \dfrac{31}{16} = 1\dfrac{15}{16}$

Sum of an infinite geometric progression(GP)

$S_\infty = \dfrac{a}{1 - r}\ \text{ (if -1 < r < 1)}$

where a= the first term , r = common ratio

Example
Find $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}+\cdots \infty$

a = 1,       r = $\dfrac{\left(\dfrac{1}{2}\right)}{1} = \dfrac{1}{2}$

Here -1 < r < 1. Hence,
$S_\infty = \dfrac{a}{1 - r} = \dfrac{1}{\left(1 - \dfrac{1}{2}\right)} = \dfrac{1}{\left(\dfrac{1}{2}\right)} = 2$

Geometric Mean

If three non-zero numbers a, b, c are in GP, b is the Geometric Mean(GM) between a and c. In this case, $b = \sqrt{ac}$

The Geometric Mean(GM) between two numbers a and b = $\sqrt{ab}$

(Note that if a and b are of opposite sign, their GM is not defined.)

To solve most of the problems related to GP, the terms of the GP can be conveniently taken as
3 terms: $\dfrac{a}{r}$, a, ar
5 terms: $\dfrac{a}{r^2}$, $\dfrac{a}{r}$, a, ar, ar2

If a, b, c are in GP, $\dfrac{a-b}{b-c} = \dfrac{a}{b}$

In a GP, product of terms equidistant from beginning and end will be constant.

Relationship Between Arithmetic Mean, Harmonic Mean, and Geometric Mean of Two NumbersIf GM, AM and HM are the Geometric Mean, Arithmetic Mean and Harmonic Mean of two positive numbers respectively, then

GM2 = AM × HM
Some Interesting Properties to Note
Three numbers a, b and c are in AP if $b = \dfrac{a + c}{2}$

Three non-zero numbers a, b and c are in HP if $b = \dfrac{2ac}{a + c}$

Three non-zero numbers a, b and c are in HP if $\dfrac{a - b}{b - c} = \dfrac{a}{c}$
Let A, G and H be the AM, GM and HM between two distinct positive numbers. Then

(1) A > G > H
(2) A, G and H are in GP
If a series is both an AP and GP, all terms of the series will be equal. In other words, it will be a constant sequence.
Power Series : Important formulas

$1 + 1 + 1 + \cdots\text{ n terms}$ $= \sum{1} = n$

$1 + 2 + 3 + \cdots + n$ $=\sum{n}= \dfrac{n(n+1)}{2}%$

$1^2 + 2^2 + 3^2 + \cdots + n^2$ $= \sum{n^2}= \dfrac{n(n+1)(2n + 1)}{6}%$

$1^3 + 2^3 + 3^3 + \cdots + n^3$ $= \sum{n^3}= \dfrac{n^2(n+1)^2}{4} = \left[\dfrac{n(n+1)}{2}\right]^2 %$

mamatha
2016-03-26 07:30:24
what is the sum of n terms in HP
kumar rohit
2016-12-18 04:06:19
In hp the sum of n term is not defined
raghav
2016-01-23 17:14:58
how to find the product of terms which are in ap and gp?
sam
2016-02-03 12:38:40
Opara
2015-11-28 08:31:48
Hi great one, thanks a million for the enlightenment.

Under the section "5. Some Interesting Properties to Note", I observed you wrote HP rather than GP here

Three non-zero numbers a, b and c are in HP if (a−b)/(b-c) = a/c
jiju (Junior Maths Expert, careerbless.com)
2015-12-23 10:14:47
It is HP only.

Suppose (a−b)/(b-c) = a/c
=> (a-b)c = (b-c)a
=> ac - bc = ab - ac
=> 2ac = ab + bc
=>  2/b = 1/c + 1/a  (divided all terms by abc)

i.e., 1/a , 1/b, 1/c are in AP
i.e.,  a,b,c are in HP
macoi
2015-06-18 12:35:06
mathematically speaking, the next term cannot be determined by giving only the first finite number of terms of a general sequence. explain this fact by giving examples
Mohinish
2016-05-03 07:56:17
You can establish a relationship between the first few terms.

For instance consider the following incomplete sequence.
2,4,8,?

You can see that each term is twice it's predecessor. So, the unknown number is twice it's predecessor(2*8=16). Unknown number is 16.
sani
2015-05-11 10:59:13
I hate sequence and series but after watching these formulas arrange in such an efficient manner i now i feel relaxed now i am not soooooooo much confuse so thanks.
pardeep
2015-03-29 09:09:19
354 180 64 21 10.2 ?
12345Next Go

Add a new comment...  (Use Discussion Board for posting new aptitude questions.)

Name:
Email: (optional)