CBSE Class 10 Maths Notes & Important Questions Chapter 5 Arithmetic Progressions

CBSE Class 10 Maths Notes & Important Questions Chapter 5 Arithmetic Progressions

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CBSE Class 10 Maths Notes & Important Questions Chapter 5 Arithmetic Progressions

CBSE Class 10 Maths Notes Chapter 5 Arithmetic Progressions

Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed constant to the preceding term. The fixed constant is called the common difference, denoted by d. The first term of the sequence is denoted by a1.

The nth term of an AP is given by: an = a1 + (n-1)d

The sum of first n terms of an AP is given by: Sn = n/2[2a1 + (n-1)d]

If we know the first term (a1), common difference (d), and number of terms (n), then we can find the nth term and sum of first n terms of an AP.

The properties of an AP are as follows:

  • The common difference is the same between any two consecutive terms.
  • The sum of any two terms equidistant from the beginning and end of the sequence is constant.
  • The average of any two terms equidistant from the beginning and end of the sequence is the same as the average of the first and last terms.

Examples of AP: 1, 3, 5, 7, 9,… -2, 1, 4, 7, 10,… 6, 13, 20, 27, 34,…

An important application of AP is in calculating simple and compound interests.

CBSE Class 10 Maths Important Questions Chapter 5 Arithmetic Progressions

  1. Find the sum of first 15 multiples of 3.
  2. In an AP, if a1 = 5 and d = 3, find the value of a10.
  3. The 4th term of an AP is 12 and the 10th term is 36. Find the common difference and the first term.
  4. Find the sum of first 25 terms of an AP if the first term is 3 and the common difference is 2.
  5. In an AP, if a3 + a7 = 24 and a5 + a9 = 38, find a1 and d.
  6. If the sum of first n terms of an AP is given by Sn = 5n^2 + 2n, find the first term and common difference.
  7. The first term of an AP is 2 and the common difference is -3. Find the sum of the first 20 terms.
  8. In an AP, the sum of the first n terms is given by Sn = 3n^2 + 2n. Find the common difference and the 10th term.
  9. Find the sum of all two-digit natural numbers which are divisible by 3.
  10. If the 4th term of an AP is 17 and the 17th term is 4 more than thrice the 4th term, find the common difference and the first term.

CBSE Class 10 Maths Important Questions Answers Chapter 5 Arithmetic Progressions

  1. The sum of first 15 multiples of 3 = 3 + 6 + 9 + … + 45 = 3(1 + 2 + 3 + … + 15) = 3(15 × 16/2) = 360

Therefore, the sum of first 15 multiples of 3 is 360.

  1. We know that the nth term of an AP is given by an = a1 + (n-1)d. Therefore, a10 = a1 + (10-1)3 = a1 + 27.

Given a1 = 5, we have a10 = 5 + 27 = 32.

Therefore, the value of a10 is 32.

  1. Let the first term of the AP be a1 and the common difference be d.

Then, a4 = a1 + 3d = 12 a10 = a1 + 9d = 36

Subtracting the first equation from the second equation, we get 6d = 24, which gives d = 4.

Substituting d = 4 in the first equation, we get a1 + 12 = 12, which gives a1 = 0.

Therefore, the common difference is 4 and the first term is 0.

  1. We know that the sum of first n terms of an AP is given by Sn = n/2[2a1 + (n-1)d]. Substituting the given values, we get:

S25 = 25/2[2(3) + (25-1)(2)] = 25/2[6 + 48] = 25 × 27 = 675

Therefore, the sum of first 25 terms of the AP is 675.

  1. We have a3 + a7 = 24 and a5 + a9 = 38.

Using the formula for the nth term of an AP, we can write a3 = a1 + 2d, a5 = a1 + 4d, a7 = a1 + 6d, and a9 = a1 + 8d.

Substituting these values in the given equations, we get:

2a1 + 14d = 24 2a1 + 28d = 38

Subtracting the first equation from the second, we get 14d = 14, which gives d = 1.

Substituting d = 1 in the first equation, we get 2a1 + 14 = 24, which gives a1 = 5.

Therefore, the first term is 5 and the common difference is 1.

  1. We know that Sn = 5n^2 + 2n. Substituting the value of Sn from the formula, we get:

5n^2 + 2n = n/2[2a1 + (n-1)d]

Simplifying this equation, we get:

10a1 + 5n^2d + 5nd – n = 0

This is a quadratic equation in n. Solving for n, we get:

n = (-5d ± sqrt(25d^2 + 40a1))/10

We need to choose the positive root, as n cannot be negative. Therefore, n = (-5d + sqrt(25d^2 + 40a1))/10.

Since the first term is a1 and the common difference is d, we can substitute these values in the above equation and solve for a1 and d.

CBSE Class 10 Maths Important Questions Answers MCQs Chapter 5 Arithmetic Progressions

If the common difference of an arithmetic progression is 3, what is the nth term of the progression if the first term is 7?
a) 3n + 4
b) 3n + 7
c) 3n + 10
d) 3n + 13
Answer: b) 3n + 7

What is the sum of first 20 terms of an arithmetic progression if the first term is 10 and the common difference is 2?
a) 810
b) 820
c) 830
d) 840
Answer: b) 820

The 10th term of an arithmetic progression is 25 and the 20th term is 35. What is the first term of the arithmetic progression?
a) 5
b) 10
c) 15
d) 20
Answer: b) 10

What is the common difference of an arithmetic progression if the sum of the first 15 terms is 195 and the first term is 5?
a) 5
b) 6
c) 7
d) 8
Answer: b) 6

What is the sum of the first 10 even numbers?
a) 110
b) 120
c) 130
d) 140
Answer: b) 120

CBSE Class 10 Maths Notes Chapter 5 Arithmetic Progressions

SEQUENCE:
A sequence is an arrangement of numbers in a definite order and according to some rule.
Example: 1, 3, 5,7,9, … is a sequence where each successive item is 2 greater than the preceding term and 1, 4, 9, 16, 25, … is a sequence where each term is the square of successive natural numbers.

TERMS :
The various numbers occurring in a sequence are called ‘terms’. Since the order of a sequence is fixed, therefore the terms are known by the position they occupy in the sequence.
Example: If the sequence is defined as
Arithmetic Progressions Class 10 Notes Maths Chapter 5 Q1.1

ARITHMETIC PROGRESSION (A.P.):
An Arithmetic progression is a special case of a sequence, where the difference between a term and its preceding term is always constant, known as common difference, i.e., d. The arithmetic progression is abbreviated as A.P.

The general form of an A.P. is
∴ a, a + d, a + 2d,… For example, 1, 9, 11, 13.., Here the common difference is 2. Hence it is an A.P.

In an A.P. with first term a and common difference d, the nth term (or the general term) is given
by .
an = a + (n – 1)d.
…where [a = first term, d = common difference, n = term number
Example: To find seventh term put n = 7
∴ a7 = a + (7 – 1)d or a7 = a + 6d

The sum of the first n terms of an A.P. is given by
Sn = n2[2a + (n – 1)d] or n2[a + 1]
where, 1 is the last term of the finite AP.

If a, b, c are in A.P. then b = a+c2 and b is called the arithmetic mean of a and c.

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