Progression and Series - Arithmetic, Geometric and Harmonic - ObjectiveBooks

Progression and Series - Arithmetic, Geometric and Harmonic

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:-
1. 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.
2. If each term of an A.P. be multiplied or divided by a nonzero constant, then the resulting terms are also in A.P.
3. Any term (except 1st and last) is equal to half of the sum of two terms equidistant from it.
4. The sum of n arithmetic means between ‘a’ and ‘b’ is [½ n (a + b)].
5. Sum or differences of corresponding terms of two A.P‘s are also in A.P.
6. 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:-
1. If each term of a G.P. be multiplied by or divided by a nonzero constant, then the resulting terms are also in G.P.
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.
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