Arithmetic, Geometric and Harmonic Progression and Series, Arithmetic, Geometric and Harmonic Mean:
Arithmetic Progression (A.P.):
A set of numbers denoted by t₁, t₂, t₃, ….., tn, … taken
in order is called a sequence which
is formed under a definite rule such that on putting n = 1, 2, 3, …. in the
n-th term tn, we get t₁, t₂,
t₃, ….. i.e., 1st, 2nd, 3rd, ….. terms of the sequence respectively.
If t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = ….. = d (say)
Then the above sequence is called an
Arithmetic Progression (A.P.) in
which the 1st term t₁ is usually
denoted by ‘a’ and ‘d’ is called the common
difference (C.D.). The A.P. may then be written as
a, a + d, a + 2d, a + 3d, ………
The n-th term being, a + (n - 1)d,
which is denoted by tn
For example,
the sequence 2, 5, 8, 11, …… is an A.P. whose 1st term is 2, common difference
is 3 and the n-th term is 2 + (n - 1)d, i.e., (3n - 1).
Of any three
consecutive terms of an A.P., the middle term is called the Arithmetic Mean (A.M.) of the two other
terms. If a, b, c be any three consecutive terms in an A.P., then the A.M
between ‘a’ and ‘c’ is ‘b’, where b = (a + c)/2
Note:-
- If a nonzero constant be added to or subtracted from each term of an A.P., then the resulting terms are also in A.P.
- If each term of an A.P. be multiplied or divided by a nonzero constant, then the resulting terms are also in A.P.
- Any term (except 1st and last) is equal to half of the sum of two terms equidistant from it.
- The sum of n arithmetic means between ‘a’ and ‘b’ is [½ n (a + b)].
- Sum or differences of corresponding terms of two A.P‘s are also in A.P.
- It is convenient to choose three numbers in A.P. as (a - 3d), (a - d), (a + d), (a + 3d) for solving problems.
Example: 01
4th term, 10th term and n-th term of an
A.P. are 7, 25 and 58 respectively. Find 1st term, C.D and the value of n.
Solution:
Let, 1st
term = a, C.D = d
Then, a + 3d
= 7
a + 9d = 25
Solving we
get, a = -2 and d = 3
Also, a + (n
- 1)d = 58
=> -2 + (n - 1)3 = 58
=> n = 21
Hence, the
values of 1st term a = -2, C.D = 7 and n = 21
Example: 02
If the sides of a right angled triangle be
in A.P., find the ratio of its sides.
Solution:
Let, the
sides be, (a - d), a, (a + d)
As the
largest side (a + d) is he hypotenuse in the triangle,
So, (a - d)²
+ a² = (a + d)²
=> a² = 4ad
=> a = 4d
Hence values
of the sides are
(a - d) =
(4d - d) = 3d
a = 4d
(a + d) =
(4d + d) = 5d
So, the
sides are 3d, 4d, 5d which are in the ratio of 3 : 4 : 5.
Sum of terms of A.P:
If the terms
t₁, t₂, t₃, ….., tn, … of an A.P. be added in the given order,
then t₁ + t₂ + t₃ + ….. + tn +
…. is called an Arithmetic Series.
If Sn be the sum of first
n terms of an A.P. a, a + d, a + 2d, ….. then,
Sn = a + (a + d) + (a + 2d) + …… + {a + (n - 2)d} + {a + (n -
1)d} …………..eq.(i)
Reversing the order of terms, we get
Sn = {a + (n - 1)d} + {a + (n - 2)d} + …… + (a + d) + a
…………………………….eq.(ii)
Adding the corresponding terms on
the right side of (i) and (ii) we get,
2Sn = n {2a + (n - 1)d}
Or, Sn = (n/2) {2a + (n -
1)d}
This is the formula for the sum of n
terms of an A.P.
Given Sn we can find tn
by using the formula tn = Sn - Sn - 1
Some Important Formulas:
1. 1 + 2 + 3 + ……… + n = [n (n +
1)]/2
2. 1 + 3 + 5 + ……… + (2n - 1) = n²
3. 2 + 4 + 6 + ……… + 2n = n (n + 1)
4. 1² + 2² + 3² + ……… + n² = [(n
+ 1) (2n + 1)]/6
5. 13
+ 23 + 33 + …….. + n3 = {n (n + 1)/2}²
Example: 03
Find the number of terms in the A.P. if -9 - 1 + 7 + …… to n terms =
2175.
Solution:
Here, 1st term, a = -9 and C.D = 8
We have, Sn = (n/2) {2a +
(n - 1)d}
So, n/2 {2 × (-9) + (n- 1) × 8} = 2175
=> 4n² - 13n - 2175 = 0
=> (n - 25) (4n + 87) = 0
=> n = 25, since n cannot be negative or
fraction.
Geometric Progression (G.P.):
A sequence t₁, t₂, t₃, ….., tn, … is called a Geometric Progression (G.P.) if
t₂/t₁ = t₃/t₂ = t₄/t₃ = ….. = tn+1/tn = ….. = r
(say)
The 1st term t₁ is usually denoted
by ‘a’ and ‘r’ is called the common
ratio (C.R.)
Hence, this G.P. may be written as:
a, ar, ar², ……, arn - 1, …….
The n-th term being arn-1 denoted
by tn
For example,
3, 9, 27, …… is a G.P. whose 1st term is 3, C.R. is 3 and the n-th term is 3. 3n
- 1, i.e. 3n
Of any three
consecutive terms of a G.P., the middle term is called the Geometric Mean (G.M) of the two other terms.
If a, b, c
be consecutive terms of a G.P., then the G.M. between a and c is, b = √(a × c)
Note:-
2. If each term of a G.P. be raised to the same positive integral power, then the resulting terms are also in G.P.
3. If a >
0, r > 1; the G.P. is said to be increasing.
If a > 0, 0 < r < 1; the G.P. is
said to be decreasing.
If a < 0, r > 1; the G.P. is said to
be decreasing.
If a < 0, 0 < r < 1; the G.P. is
said to be increasing.
4. If three
numbers are in G.P., we choose them as a/r, a, ar and if four numbers are in
G.P., we choose them as a/r3, a/r, ar, ar3 for
convenience in solving problems.
Example: 04
The A.M and the G.M. between ‘a’ and ‘b’
are 34 and 16 respectively. Find ‘a’ and ‘b’.
Solution:
We have, (a
+ b)/2 = 34
=> (a + b) = 17
And, √ab = 16
=> ab = 256
Solving, a =
4, b = 4
Example: 05
Three numbers are in A.P. and their sum is
21. If 1, 2, 15 are added to them in order, then the resulting numbers are in
G.P. Find the given numbers.
Solution:
Let the
numbers be, a -d, a, a +d
Then, (a -d)
+ a + (a +d) = 21
=> a = 7
So the
numbers are, (7 - d), 7, (7 + d).
Since, (7 -
d + 1), (7 + 2), (7 +d + 15), i.e. (8 - d), 9, (22 + d) are in G.P
So, (8 - d)
(22 + d) = 9²
=> d = 5 or -19
If d = 5,
the given numbers are 2, 7, 12 and
If d = -19,
the given numbers are 26, 7, -12.
Sum of terms of G.P.
If the terms
t₁, t₂, t₃, ….., tn, …of a G.P. be added in the given order, then t₁ + t₂ + t₃ + ….., tn + … is
called a Geometric Series.
Let, Sn
be the sum of first n terms of the G.P. a, ar, ar², ……
Then, Sn
= a + ar + ar² + …… + arn - 2 + arn - 1 ……………………….eq.(i)
Multiplying
both sides by r,
rSn = ar + ar² + ar3
+ …… + arn - 1 + arn ……………….………eq.(ii)
(i) - (ii)
=> Sn - rSn
= a - arn
=> Sn = {a (1 - rn)}/(1
- r), r ≠ 1
This is the
formula for the sum of first n terms of a G.P.
Example: 06
Find the first n terms of a geometric
series whose 3rd term is 4 and 9th term is 32.
Solution:
Let, 1st
term = a and C.R. = r
Given, ar² =
4 and
ar8 = 32
Solving we
get, a = 2 and r = √2
So, Sn
= {a (rn - 1)}/(r - 1) = [2 (√2n - 1)] / (√2 - 1)
= 2 (√2 + 1) (√2n -1).
Example: 07
Find the sum of first n terms of .7 + .77 +
.777 + ………
Solution:
Sn
= .7 + .77 + .777 + …. to n terms
= 7/9 (.9 + .99 + .999 + to n terms)
= 7/9 [(1 - .1) + (1 - .01) + (1 - .001) +
…to n terms]
= [7/9 (1 + 1 + 1 + …. to n terms)] - [7/9
(.1 + .01 + .001 + …. to n terms]
= [7n/9] - [(7/9) {.1(1 - 1n)/(1
- .1)}]
= [7n/9] - [7/81) [1 - {1/10n)}]
Harmonic Progression (H.P.):
A sequence
of terms whose reciprocals are in A.P. is called a Harmonic Progression (H.P.)
e.g.,
corresponding to an A.P. 1, 2, 3, 4, ….the H.P.is 1, 1/2, 1/3, 1/4, ……
Corresponding
to an H.P., we always have an A.P. and vice versa. In most of the cases, there
is no general method for solving problems related to H.P. and that is why, the
H.p. related problems are usually solved by writing down the corresponding A.P.
Of three
consecutive terms in an H.P., the middle term is called the Harmonic Mean (H.M.) of the two other
terms. If a, b, c be three consecutive terms in an H.P. then, 1/a, 1/b/ 1/c are
in A.P. and so
1/b = [1/a + 1/c] / 2
Hence, the
H.M. between ‘a’ and ‘c’ is 2ac/(a + c).
Note:-
If each term
of an H.P. be multiplied or divided by a nonzero constant, then the resulting
terms are also in H.P.
Example: 08
Write down the H.P. whose 4th term is 1/7
and 13th term is 1/34.
Solution:
Let ‘a’ be
the first term and ‘d’ be the common difference of the corresponding A.P.
Then, a + 3d
= 7 and a + 12d = 34
Solving, we
get, a = -2 and d = 3
So, the A.P.
is -2, 1, 4, 7, …….
Hence, the
H.P. is -1/2, 1, 1/4, 1/7, ……
Relation between A.M., G.M. and H.M.
If A, G, H
be the A.M., G.M. and H.M. between two distinct positive real numbers.
Then, A = (a
+ b)/2
G = √ab
H = 2ab/(a + b)
∴ [(a + b)/2]
[2ab/(a + b)] = ab = G²
Or, AH = G²
Again, A - G
= (a + b)/2 - √ab = (a + b - 2√ab)/2 = (√a² - √b²)/2
∴ A - G > 0
=> A > G
=> A/G > 1 => G/H > 1 => G > H
Hence, A > G > H
Note: If a =
b, then A = G = H.