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.
Formula
An arithmetic progression is given by a, (a + d), (a + 2d), (a + 3d), ...
where a = the first term , d = the common difference
Example
1, 3, 5, 7, .....?
Arithmetic progression (AP) with a = 1 and d = 2
a, (a + d), (a + 2d), (a + 3d),(a + 4d)........
1, 3, 5, 7, (1+4(2))
1, 3, 5, 7, 9
Ans : 9
Exercise
Arithmetic progression (AP)
a = 7 and d= 6
a, (a + d), (a + 2d), (a + 3d),(a + 4d)........
7, 13, 19, 25, (7+4(6))
7, 13, 19, 25,31
n$^{th}$ term of an arithmetic progression
Formula
tn = a + (n – 1)d
where t$_{n}$ = n$^{th}$ term, a= the first term , d= common difference
Example
Find 10$^{th}$ term in the series 1, 3, 5, 7, ...
a = 1
d = 3 – 1 = 2
10$^{th}$ term, t$_{10}$ = a + (n-1)d = 1 + (10 – 1)2 = 1 + 18 = 19
Exercise
a = 7
d = 13 – 7 = 6
n=16
16$^{th}$term, t$_{16}$ = a + (n-1)d = 7 + (16 – 1)6 = 7 + 90 = 97
16$^{th}$term = 97
Number of terms of an arithmetic progression
Formula
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 = $\dfrac{(l−a)}{d} + 1$ = n = $\dfrac{(72−8)}{4} + 1$
=n = $\dfrac{(64)}{4} + 1$ = 16+1 = 17
Sum of first n terms in an arithmetic progression
Sn= $\dfrac{n}{2}$ [ 2a+(n−1)d ] = $\dfrac{n}{2}$ (a+l)
where a = the first term,
d= common difference,
l = t$_n$ = n$^{th}$ term = a + (n-1)d
Example
Find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms
a = 4
d = 7 – 4 = 3
Sum of first 20 terms, S$_{20}$
= $\dfrac{n}{2}$ [ 2a+(n−1)d ]
= $\dfrac{20}{2}$ [ (2 * 4) + (20−1)3 ]
= 10[8+(19×3)]
= 10(8+57)
= 650
Exercise
a = 6
l = 30
d = 9 – 6 = 3
n = $n = \dfrac{(l - a)}{d} + 1$
= $\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 = $\dfrac{1}{2}(a + c)$
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)
$T_n$ = $S_n$ - $S_n-1$
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 $\dfrac{1}{a_1}, \dfrac{1}{a_2},\dfrac{1}{a_3}, ..... \dfrac{1}{a_n}$ are in AP. Harmonic Progression is also known as harmonic sequence.
Example
$\dfrac{1}{2}, \dfrac{1}{6}, \dfrac{1}{10},$...... is a harmonic progression (HP)
If a, (a+d), (a+2d), . . . are in AP, nthterm of the AP = a + (n - 1)d
Hence, if $\dfrac{1}{a}, \dfrac{1}{a+d}, \dfrac{1}{a+2d},$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.
Formula
A geometric progression(GP) is given by a, ar, ar2, ar3, ...
where a = the first term , r = the common ratio
Example
1, 3, 9, 27, ... is a geometric progression(GP) with a = 1 and r = 3
Exercise
nth term of a geometric progression(GP)
Formula
$t_n = ar^{n-1}$
where tn = nth term, a= the first term , r = common ratio, n = number of terms
Example
Find the 10th term in the series 2, 4, 8, 16, ...
a = 2, r = $\dfrac{4}{2}$ , n = 10
n = 10
10th term, t10
=$ar^{n-1} = 2 \times 2^{10-1}$
= $2 \times 2^{9} = 2 \times 512 = 1024 $
Exercise
a = 5, r - $\dfrac{15}{5}$ = 3, n=5
5th term, t5
=$ar^{n-1} = 5 \times 3^{5-1}$
= $5 \times 3^4 = 5 \times 81$ = 405
Sum of first n terms in a Geometric progression(GP)
Formula
$S_n =\begin{cases} \dfrac{a(r^n - 1)}{r - 1}\ & (\text{if } r \gt 1) \\ \\ \dfrac{a(1 - r^n)}{1 - r}\ & (\text{if } r \lt 1) \\\end{cases}$
where a= the first term,
r = common ratio,
n = number of terms
Example
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} $= $2 \times 28$
= 1456
Exercise
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)
Formula
$S_\infty$=$ \dfrac{a}{1 - r}$ (if -1 < r < 1)
where a= the first term , r = common ratio
Example
Find 1 + $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}+.....\infty$
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.)
Additional Notes on GP
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 Numbers
If GM, AM and HM are the Geometric Mean, Arithmetic Mean and Harmonic Mean of two positive numbers respectively, then
GM$^2$ = 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+⋯ n terms = ∑1=n
1+2+3+⋯+n =∑n = $\dfrac{n(n+1)}{2}$
$1^2 + 2^2 + 3^2 + ........ +n^2 = ∑ n^2$ = $\dfrac{n(n+1)(2n + 1)}{6}$
$1^3 + 2^3 + 3^3 + ....... + n^3 = ∑ n^3$ = $\dfrac{n^2(n+1)^2}{4}$ = $\left[\dfrac{n(n+1)}{2}\right]^2 $