AP Class 10 Maths Chapter 6 Progressions

The AP Class 10 Maths Chapter 6 Progressions discusses patterns in which succeeding terms are obtained by adding or multiplying preceding terms by a fixed number. We also learn how to find the nth term of a progression and the sum of n consecutive terms.

What is an Arithmetic Progression?

A list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term. The fixed number d is called the common difference.

The terms of an AP is represented as \(a, a+d, a+2d, a+3d….\)

  • A given list of numbers \(a_{1},a_{2},a_{3}….\) is said to be an arithmetic progression if \(a_2-a_1, a_3-a_4\) and so on give the same number.
  • The nth term in an Arithmetic Progression can be determined using the first term a and the common difference d as follows:
  • \(a_n=a+(n-1)d\)

  • The sum of the n terms of an Arithmetic Progression is given by the formula
  • \(S=\frac{n}{2}[2a+(n-1)d]\)

  • If a given Arithmetic Progression is finite with the last term l, the sum of all the terms in an arithmetic progression is computed using the formula
  • \(S=\frac{n}{2}(a+l)\)

What is a Geometric Progression?

A geometric progression is a list of numbers in which each term is obtained in which each term is obtained by multiplying the preceding term by a fixed number ‘r‘ except the first term. The general form of a geometric progression is \(a,ar,ar^{2},ar^{3}\). The nth term in a geometric progression can be obtained if the first term and common ratio of the GP are known as follows \(a_{n}=ar^{n-1}\).

Class 10 Maths Chapter 6 Progressions Chapter Questions

  1. Find the sum of first 30 positive integers divisible by 5
  2. Solution:

    The first 30positive integers divisible by 5 are 5,10,15,20….

    Since the difference between each term is the same, it qualifies as an arithmetic progression.

    To find the sum of the first 30 positive integers divisible by 5, we can use the formula

    \(S=\frac{n}{2}[2a+(n-1)d]\)

    Here, a = 5, d = 5 and n = 30

    Substituting the values in the equation, we get

    \(S=\frac{30}{2}[(2\times 5)+(30-1)\times 5]\) \(S=15[10+145]\) \(S=15\times 145=2175\)

    S = 2175

    The sum of the first 30 positive integers divisible by 5 is 2175.

  1. Find the 12th term of a GP whose 8th term is 192 and the common ratio is 2
  2. Solution:

    Let us find the first term a using the formula

    \(a_{n}=ar^{n-1}\)

    We know that the r = 2 and \(a_8=192\), substituting the values in the

    equation, we get

    \(192=a\times 2^{8-1}\) \(a=\frac{192}{2^{7}}\)

    Now to find the 12th term, we use the formula

    \(a_{n}=ar^{n-1}\)

    Substituting n =12, r = 2 and \(a=\frac{192}{2^{7}}\), we get

    \(a_{12}=\frac{192}{2^{7}}\times 2^{(12-1)}\) \(192\times 2^{(11-7)}\)=3072

    The 12th term is 3072.

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Practise This Question

Which of the following sequences form an AP?

(i) 2, 4, 8, 16…….. 

(ii) 2, 3, 5, 7, 11…….

(iii) -1, -1.25, -1.5, -1.75……

(iv) 1, -1, -3, -5, -7…………