Arithmetic Progression Maths Notes

Quadratic Equations Maths Notes

Sequence

  1. 1, 2, 3, 4, 5, … are serial numbers in a set of natural numbers.
  2. 1, 3, 5, 7, 9, … are odd numbers in a set of natural numbers.
  3. 1, 4, 9, 16, 25, … are the squares of the serial numbers in a set of natural numbers.

In each of above arrangements, there is some relationship between any two consecutive numbers.
A sequence is a collection of numbers arranged in a definite order according to some definite rule.
The numbers are denoted by a1, a2, a3, ……, an. a1 is the first term, a2 is the second term, …, an is the nth term.
The sequence of numbers is denoted by f1, f2, f3

Arithmetic Progression Maths Notes

Complete the following pattern:

1.
Arithmetic Progression 1

2.
Arithmetic Progression 2
The formula for the first pattern : 2n – 1, where n = 1, 2, 3, ….
The formula for the second pattern : 3n + 2, where n = 1, 2, 3, …….

Terms in a sequence

The terms in serial order f a sequence are denoted by t1, t2, t3, …, tn.
A sequence is usually denoted by { tn }.

Arithmetic Progression Maths Notes

Arithmetic Progression

Some sequences are given below. For every sequence write the next two terms:

  1. 100, 70, 40, 10, ……
  2. -7, -4, -1, 2, ……
  3. 4, 4, 4, ……

In the given sequences, observe how the next term is obtained.
Arithmetic Progression 3
In each sequence above, every term is obtained by adding a particular number to the previous term. The difference between two consecutive terms is constant.

The difference in ex. (j) is negative, in ex. (u) it is positive and in ex. (iii) it is zero. If the difference between two consecutive terms is constant, then it is called the common difference and is generally denoted by letter d.

Remember:

In the given sequence, if the difference between two consecutive terms (tn + 1 – tn) is constant, then the sequence is called Arithmetic Progression (A.P.). In this sequence tn + 1 – tn = d is the common difference.

In an A.P., if the first term is denoted by a and the common difference is d then,
t1 = a, t2 = a + d, t3 = (a + d) + d = a + 2d. i.e. the next term is obtained by adding d to the previous term.
A.P. having first term as a and common difference d is a, (a + d), (a + 2d), (a + 3d), ………….

[Note : (1) The value of d may be positive, negative or zero.
(2) If tn+1 – tn is constant, then for every n ∈ N, the sequence is an A.P.
(3) In an A.P., t n + 1 – tn = tn – tn – 1 = d.
(4) The set of even / odd numbers is an A.P.]

Arithmetic Progression Maths Notes

Let’s think:

Is 5, 8, 11, 14, … an A.P.? If so, then what wifi be the 100th term?

Check whether 92 is in this A.P. Is the number 61 in this A.P.?

  • This is an AY., because the common difference is 3 which is constant.
  • The 100th term will be 302.
  • 92 is the 30th term of this A.P.
  • The number 61 is not in this A.P.

[Note : The answers to the questions (ii) to (iv) can be given only after learning the formula for the nth term of an A.P.]

nth Term of an A.P.

Determination of the formula for the nth term:
Generally, in the AP. t1, t2, t3, ……, if the first term
is a and the common difference is d, then
t1 = a
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
∴ we get the formula tn = a + (n – 1)d
[Note: For an A.P., if d = 0, the sequence is a constant one.]

Arithmetic Progression Maths Notes

Let’s think

Kabir’s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, “it seems like Kabir’s height grows in Arithmetic Progression”. Assuming this, she calculated how tail Kabir will be at the age of 15 years when he is in the 10th ! She was shocked to find it. You to assume that Kabir’s height grows in A.P. and find out his height at the age of 15 years.

Here, t1 = a = 70 cm, t2 = 80 cm; t3 = 90 cm.
d = t3 – t2 = 90 – 80 = 10cm
We want to find t15
tn = a + (n – 1)d … (Formula)
∴ t15 = 70 + (15 – 1) × 10 … (Substituting the values)
= 70 + 14 × 10 = 70 + 140 = 210 cm
So, Kabir’s height when he will be in 10th will be 210 cm ! Wow.

Let’s learn

1. Derive the formula for the nth term of the sequence of odd natural numbers and even natural numbers. The difference between any two consecutive odd or even natural numbers is 2, i.e. d = 2.
For odd natural numbers:
1, 3, 5, 7, …
tn = a + (n – 1)d
= 1 + (n – 1) × 2
= 1 + 2n – 2 = 2n – 1
The formula for nth term of the sequence of odd natural numbers : tn = 2n – 1.
For even natural numbers:
2, 4, 6, 8, …
tn = a + (n – 1)d
= 2 + (n – 1) × 2
= 2 + 2n – 2 = 2n
The formula for nth term of the sequence of even natural numbers: tn = 2n.

Arithmetic Progression Maths Notes

2. Find the sum of the first n (i) natural numbers
(ii) odd natural numbers and
(iii) even natural numbers.
Answer:
(i) For first n natural numbers:
Here, a = 1, d = 1, nth term = n
Arithmetic Progression 4

(ii) For odd natural numbers:
Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{n}{2}\)[2 × 1 + (n – 1) × 2]
= \(\frac{n}{2}\) [2 + 2n – 2]
= \(\frac{n}{2}\)(2n) = n2
∴ Sn = n2.

(iii) For even natural numbers:
S = \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{n}{2}\) [2 × 2 + (n – 1) × 2]
= \(\frac{n}{2}\) [4 + 2n – 2]
= \(\frac{n}{2}\) (2n + 2) = n(n + 1)
∴ S = n(n + 1).

Arithmetic Progression Maths Notes

Application of A.P.

We apply the theory and formulae of an A.R in solving various word problems which we come across in day-to-day life.

Maths Notes

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