Arithmetic Progressions

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Framing the rule
Ex 1: 4, 6, 8, 10, . . . . In above: 4, 6, 8, 10 . . . . are called terms of the sequence. Clearly, we observe that each term is obtained by adding 2 to the previous term. In this sequence, the rule is: to get a term, add 2 to the previous term. Ex 2: 2, 6, 18, 54, . . . . In above: 2, 6, 18, 54. . . . are called terms of the sequence. Clearly, we observe that each term is obtained by multiplying the previous term with 3 In this sequence, the rule is: multiply a term by 3 to get the next term. Ex 3: 12, 6, 0, – 6, – 12 . . . . In the above sequence, the first term = 12 Second term = 12 – 6 = 6 Third term = 6 – 6 = 0 Fourth term = 0 – 6 = – 6 Clearly, we observe that each term is obtained by subtracting 6 from the previous term. In this sequence, the rule is: subtract 6 from the previous term. Ex 4: 4, 9, 16, 25, 36 This is a finite sequence and is equal to 2², 3², 4², 5², 6² In this sequence, the terms are squares of the natural numbers from n = 2 to n = 6.

Ex 1: 4, 6, 8, 10, . . . .
In above: 4, 6, 8, 10 . . . . are called terms of the sequence.
Clearly, we observe that each term is obtained by adding 2 to the previous term.
In this sequence, the rule is: to get a term, add 2 to the previous term.

Ex 2: 2, 6, 18, 54, . . . .

In above: 2, 6, 18, 54. . . . are called terms of the sequence.
Clearly, we observe that each term is obtained by multiplying the previous term with 3
In this sequence, the rule is: multiply a term by 3 to get the next term.

Ex 3: 12, 6, 0, – 6, – 12 . . . .

In the above sequence, the first term = 12
Second term = 12 – 6 = 6
Third term = 6 – 6 = 0
Fourth term = 0 – 6 = – 6
Clearly, we observe that each term is obtained by subtracting 6 from the previous term.
In this sequence, the rule is: subtract 6 from the previous term.

Ex 4: 4, 9, 16, 25, 36
This is a finite sequence and is equal to
2², 3², 4², 5², 6²
In this sequence, the terms are squares of the natural numbers from n = 2 to n = 6.

Sequence and Series

Sequence:

A set of numbers arranged in a definite order according to some definite rule is called a sequence. Each number of the set is called a term of the sequence.

Finite and Infinite sequence:
A sequence is called finite or infinite according to the number of terms in it is finite or infinite.

General term of sequence:
Let us consider the sequence of "cubes" of natural numbers: 1, 8, 27, 64, . . . .
The different terms of a sequence are usually denoted by t₁, t₂, t₃, . . . etc.
Here, the subscript (always a natural number) denotes the position of the term in the sequence.
Thus, in the above sequence t₁ = 1; t₂ = 8; t₃ = 27 . . . . . etc.
Hence, First term = t₁ = 1, Second term = t₂ = 8, Third term = t₃ = 27, . . . etc.
In general, n^th term = t_n, which is called general term of the sequence.

Often, it is possible to express the rule which generates the various terms of the sequence in terms of an algebraic formula.
In the above sequence 1, 8, 27, 64, 125 . . ., n^th term = t_n = n³.
Thus, the rule for the above sequence is n³, where n is any natural number.

Series:

If a₁, a₂, . . ., a_n is a sequence of numbers,
then the expression a₁ + a₂ + . . . + a_n
is called series associated with the given sequence.
Like a sequence, a series may also be finite or infinite.

It is common to represent a series compactly using the Σ (sigma) symbol. Sigma indicates a summation as:

a_i = a₁ + a₂ + a₃ + . . . + a_n for a finite series.

Note that 'i' takes the values from '1' to the number indicated at the top of the sigma symbol.

(The L.H.S. is read as "sigma a_i, i equal to 1 to n")

a_i = a₁ + a₂ + a₃ + . . . + a_n + . . . ∞ for an infinite series.

Example

Example
If the 10^th term of an A.P. is 50 and the 17^th term is 20 more than the 13^th term, then find A.P. Sol: Let a be the first term and d be the common difference. Then, t₁₀ = a + 9d t₁₃ = a + 12d t₁₇ = a + 16d Given, t₁₀ = 50 t₁₇ = t₁₃ + 20 a + 9d = 50 ---- (i) t₁₇ = t₁₃ + 20 ⇒ a + 16d = a + 12d + 20 ⇒ d = 5 Substitute the ‘d’ value in equation (i) a + 9d = 50 ⇒ a + 9(5) = 50 ⇒ a = 5 ∴ Finally, a = 5, d = 5 Now, we have to find A.P. We know that if a is the first term and d is the common difference, then the arithmetic progression is a, a + d, a + 2d, a + 3d . . . . ∴ The required A.P. is = 5, [5 + 5], [5 + 2(5)], [5 + 3(5)], [5 + 4(5)], . . . = 5, 10, 15, 20, 25, . . .

If the 10^th term of an A.P. is 50 and the 17^th term is 20 more than the 13^th term, then find A.P.
Sol: Let a be the first term and d be the common difference. Then,
t₁₀ = a + 9d
t₁₃ = a + 12d
t₁₇ = a + 16d
Given, t₁₀ = 50
t₁₇ = t₁₃ + 20
a + 9d = 50 ---- (i)
t₁₇ = t₁₃ + 20
⇒ a + 16d = a + 12d + 20
⇒ d = 5
Substitute the ‘d’ value in equation (i)
a + 9d = 50
⇒ a + 9(5) = 50
⇒ a = 5
∴ Finally, a = 5, d = 5
Now, we have to find A.P.
We know that if a is the first term and d is the common difference,
then the arithmetic progression is
a, a + d, a + 2d, a + 3d . . . .
∴ The required A.P. is
= 5, [5 + 5], [5 + 2(5)], [5 + 3(5)], [5 + 4(5)], . . .
= 5, 10, 15, 20, 25, . . .

Arithmetic progression

An arithmetic progression is a sequence in which each term (except the first term) is obtained by adding a fixed number (positive or negative or zero) to the term immediately preceding it.
Hence, this fixed number becomes the difference of two successive terms.
For this reason it is called as the common difference and is usually denoted by d.

Quantities are said to be in Arithmetic Progression (A.P.) when they increase or decrease by a common difference. The common difference is formed by subtracting any term of the sequence from that which follows it.

Thus, if t₁, t₂, t₃, . . . . . , t_n are the terms in an A.P. and the common difference is 'd', then
t₂ = t₁ + d d = t₂ – t₁ t₃ = t₂ + d d = t₃ – t₂ t₄ = t₃ + d d = t₄ – t₃
- - - - - - - - - - - - - - - - t_n = t_{n - 1} + d d = t_n – t_{n - 1}
In above, if the first term t₁ = a, then
t₂ = t₁ + d = a + d = a + (2 – 1)d
t₃ = t₂ + d = [a + d] + d = a + 2d = a + (3 – 1)d
t₄ = t₃ + d = [a + 2d] + d = a + 3d = a + (4 – 1)d
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
t_n = t_{n – 1} + d = a + (n – 1)d t_n is called general term of A.P. Thus t_n = a + (n – 1)d
By substituting n = 1, 2, 3, . . . we get a, a + d, a + 2d, a + 3d, a + 4d, . . . . . . . . It represents an arithmetic progression where "a" is the first term and "d" is the common difference. This is called the general form of an A.P. The common difference (d) is given by t_n – t_{n – 1}.

Example

Insertion of AMs
Insert four arithmetic means between 2 and -18. What is the common difference of the sequence ?
Sol: Let the four arithmetic means be x₁, x₂, x₃, x₄ and the common difference be d. Then 2, x₁, x₂, x₃, x₄, - 18 are in A.P. 2 is the first term and -18 is the 6^th term. i.e, a = 2, t_n = -18 and n = 6 We know t_n = a + (n – 1)d ⇒ -18 = 2 + 5d ⇒ d = -4 Hence, x₁ = 2 + 1.d = 2 + (-4) = -2 x₂ = 2 + 2.d = 2 - 8 = -6 x₃ = 2 + 3.d = 2 - 12 = -10 x₄ = 2 + 4.d = 2 - 16 = -14 ∴ The required sequence is 2, -2, -6, -10, -14, -18 with a common difference of -4.

Insertion of AMs

Insert four arithmetic means between 2 and -18. What is the common difference of the sequence ?

Sol: Let the four arithmetic means be x₁, x₂, x₃, x₄ and the common difference be d.
Then 2, x₁, x₂, x₃, x₄, - 18 are in A.P.
2 is the first term and -18 is the 6^th term.
i.e, a = 2, t_n = -18 and n = 6
We know t_n = a + (n – 1)d
⇒ -18 = 2 + 5d
⇒ d = -4
Hence,
x₁ = 2 + 1.d = 2 + (-4) = -2
x₂ = 2 + 2.d = 2 - 8 = -6
x₃ = 2 + 3.d = 2 - 12 = -10
x₄ = 2 + 4.d = 2 - 16 = -14
∴ The required sequence is
2, -2, -6, -10, -14, -18
with a common difference of -4.

Arithmetic mean

When three quantities are in A.P., the middle one is said to be the arithmetic mean of the other two.
Thus, if a, b, c are three successive quantities in A.P., then "b" is called the arithmetic mean of a and c.
It is given by b =
Hence, arithmetic mean of two numbers is half of their sum i.e, their average.
(It may be noted that, while the AM used in A.P. is average of just two numbers, the term mean used in statistics is the average of entire set of data).

Properties of Arithmetic Progression:

1. If a, b, c are in A.P., then 2b = a + c
2. If a, b, c are in A.P. and k ≠ 0, then the following are also in A.P.
(i) a + k, b + k, c + k i.e, when k is added to each term
(ii) a – k, b – k, c – k i.e, when k is subtracted from each term
(iii) ak, bk, ck i.e, when each term is multiplied by k
(iv) a/k, b/k, c/k i.e, when each term is divided by k

The innovative school boy

When Carl Friderich Gauss, the great German Mathematician, was in school, his teacher gave a problem of finding the sum of first 100 natural numbers. He immediately worked out and replied that the sum is 5050.
He wrote the first 100 natural numbers as given below:

S	=	1 + 2 + . . . . + 99 + 100
Reversing the order of the numbers at the right
S	=	100 + 99 + . . . . + 2 + 1
Adding these two, he obtained
2S	=	101 + 101 + . . . + 101 + 101 (100 times)
	=	101 * 100
S	=	101 * (100/2)
	=	101 * 50
	=	5050

Sum of n terms of an Arithmetic Progression

Refer to an actual incident explained at the right.
We will now use the same technique to find the sum of the first "n" terms of an A.P.
Let the first term of an A.P. be a and common difference be d.
Let S_n denote the sum of the n terms of A.P. Then
S_n = a + [a + d] + [a + 2d] + . . . . + [a + (n – 2)d] + [a + (n – 1)d]
Rewriting the terms in reverse order, we have
S_n = [a + (n – 1)d] + [a + (n – 2)d] + . . . . + [a + 2d] + [a + d] + a
On adding the two
S_n + S_n = [2a + (n – 1)d] + [2a + (n – 1)d] + . . . . . . .+ [2a + (n – 1)d] + [2a + (n – 1)d] (n times)
⇒ 2S_n = n[2a + (n – 1)d]
⇒ S_n = [2a + (n – 1)d]
We can also write
S_n = [a + a + (n – 1)d] = [a + a_n] since a + (n – 1)d = a_n
Sometimes S_n is simply denoted by S.
If there are only 'n' terms in an A.P. and a_n = l (the last term), then
S = [a + l]
The above formula is useful when the first and last terms of the A.P. are known.

In solving problems related to Arithmetic progressions, the following shall be useful:
i. 3 successive terms of an A.P. can be considered as: (a – d), a, (a + d)
ii. 4 successive terms of an A.P. can be considered as: (a – 3d), (a – d), (a + d), (a + 3d)
iii. 5 successive terms of an A.P. can be taken as: (a – 2d), (a – d), a, (a + d), (a + 2d)
iv. The n^th term of an A.P. is equal to sum of n terms minus sum of (n – 1) terms
i.e, T_n = S_n – S_n-1
v. The sum of terms equidistant from beginning and end is constant

A ladder with sloping sides

The lengths of rungs will be in A.P.

Real life examples of A.P.

An Arithmetic progression involves the basic principle of counting.

a. If you deposit the same amount of money every week in a piggy bank, the weekly total amounts form an A.P.

b. In an aluminum ladder with sloping sides, the lengths of each rung would form an A.P. (See adjacent figure)

c. If you hire an auto or a taxi, you will be charged a fixed amount (say Rs. x) for a certain distance. Thereafter, you will have to pay additional charges per every km travelled (say Rs. y).
So, the total fares at the end of every km form an A.P. as:
x, x + y, x + 2y, x + 3y, etc

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