Quick Recap: Arithmetic Progression (AP) & Geometric Progression (GP) in Maths for ICSE Class 10 Board Exams

Arithmetic Progression

First the definition: An Arithmetic Progression or Arithmetic Sequence is a sequence of numbers such that the difference between two successive numbers is constant.

The difference between the 1st and the 2nd term is equal to the difference between the 3rd and the 4th term or the 8th and 9th term or any other successive terms in the sequence.

Let’s look at some examples:

1, 3, 5, 7, 9…

3, 6, 9, 12, 15…

10, 20, 30, 40, 50…

Each number in the sequence is called a member/term/element.

The first or initial member of a sequence is represented by the letter a.

Common Difference, ‘d’:

The constant difference between two successive numbers is denoted by the letter ‘d’.

Ex: 2, 5, 8, 11, 14… is an arithmetic progression with the common difference ‘d’ of 3.

d = 5 – 2 = 3

Therefore, the difference between 2 and 5 can be represented by ‘d’ and this d is constant for any two successive terms in the sequence.

This ‘d’ can be positive or negative.

Positive common difference: 3, 6, 9, 12, 15… with a difference of 3

Negative common difference: 5, 3, 1, -1, -3, -5… with a difference of -3

General Term ‘n’:

The general term of an arithmetic progression is represented by the letter ‘n’ .

Ex: In the sequence 2, 5, 8, 11, 14, a=2 and d=3. The ‘n’th member would be 2+(n-1)3

Arithmetic Series ‘Sₙ’

First the definition: Arithmetic Series is the sum of the first n numbers of a sequence and is represented by Sₙ.

In the sequence of 1, 3, 5, 7, 9, the sum of the given numbers is 1+3+5+7+9=25 where l is the last term, if given. OR if the last term is not given.

A finite portion of an arithmetic series is called a Finite Arithmetic Progression. Easy, right?

Let us apply all of the above on an example. Taking the example of the finite Arithmetic Sequence: 1, 7, 13, 19, 25.

The first step is to note down all the variables we will need to solve the question in a corner:

  • Initial letter ‘a’
  • Common difference ‘d’
  • General term ‘n’ or aₙ

In this sequence, the 1st element is the number 1. Therefore a = 1.

 

Let us find out the pattern in which this progression has been set, i.e., the common difference ‘d’.

Take any 2 consecutive numbers and subtract the former from the latter. For example, let us take the 3rd and the 4th. Subtract the 3rd number from the 4th = 19-13=6. Therefore, d=6

This is the same as the formula d = a₂ – a.

Up next, the general term ‘n’ can be any number in the sequence. Sometimes, if a sequence is too long, it is difficult to find the 9th or 15th number or so, using just the general formula.

Therefore, we use the aₙ = a + (n – 1) d formula.

We’ll understand how this formula works by applying this to the above sum.

 

Let us find a₅ in this sequence. As we can see, the 5th element here is 25. Do you think we can get the same answer by applying the formula?

a₅ = 1+(5-1)6

= 1+4×6

= 1+24

a₅ = 25

 

Now, do you think you can find out the 10th element in this sequence using the same formula? Go ahead and try it.

Arithmetic Series, as we have seen, is the sum of the first ‘n’ numbers. Let us try to find the sum of the above given progression:

n=5, d=6, a=1, l=25

S₅ = 5/2 (1+25)

= 2.5 (26)

= 65

Is that correct? To check, try adding all the numbers of the AP yourself and see if you get the same answer.

Geometric Progression

Also known as Geometric Sequence, it is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed number which is not a zero.

Like in an arithmetic progression, the first term is represented by the letter ‘a’ and the fixed number, called the ratio, is represented by ‘r’.

Therefore, the sequence goes like this: a, ar, ar², ar³…arⁿ⁻¹

For example, a series with the initial term a=2 and the ratio r=3 would be:

2, 2×3, 2×3², 2×3³…which adds up to 2, 2×3, 2×9, 2×27…ultimately giving the sequence of 2, 6, 18, 54…

Finding the Common Ratio:

The Common Ratio ‘r’ is the constant number that is multiplied with each term to get the next term in the sequence. In other words, when you multiply any number in the sequence with ‘r’, you get the next term.

So, how do you find out what the value of ‘r’ is when a sequence is given?

Dividing the latter of 2 consecutive numbers by the former gives you the value of ‘r’.

Hence, r = a2 / a1.

General Term ‘aₙ’

The nth element in a sequence is found using the formula aₙ =a x r⁽ⁿ⁻¹⁾.

You can also multiply an element by the ratio to get the next element in the sequence: a₄ = a₃ x r

Geometric Series

A Geometric Series is the sum of the elements in a geometric progression. Therefore, Sₙ is the sum of the first ‘n’ elements in the sequence.

S₅ = a + (axr) + (axr²) + (axr³) + (axr⁴)

The easy formula used to find out the sum of ‘n’ elements in a sequence is:

Sₙ= [a (1 – rn)] / [1 – r]

Where a is the first term of the sequence, r is the common ratio and n is the number of elements whose sum is to be found.

 

Let us apply all the formula given above on a sample sequence.

If the initial term ‘a’ is 1 and the common ratio ‘r’ is 3, how can we make a sequence?

The sequence goes a, ar, ar², ar³…. So,

1, 1×3, 1×3², 1×3³… will be the sequence

So our geometric progression sequence is 1, 3, 9, 27…

 

Now that we have a sequence, let us pretend we do not know the ratio. Can we find it out with the formula, ?

In the sequence, a₂=3 and a₁=1.

Therefore, r = 3/1 = 3

 

Now in this sequence, we know the first 4 numbers. Can we find out the 6th term?

As we know, aₙ=axr⁽ⁿ⁻¹⁾. So if we want to find the 6th term, ‘a₆’ = axr⁽⁶⁻¹⁾

Hence, a₆=1×3⁽⁶⁻¹⁾

=1×3⁵

=243

Therefore, the 6th term of the sequence is 243. Cross check that, by multiplying the elements by the ratio till we reach the 6th term.

Some Practice Questions

Now that you have understood the concepts, try solving the following questions on your own:

Arithmetic Progression:

a) Is this an arithmetic progression? Answer with Yes or No:

  • 1, 3, 5, 7, 9
  • 2, 3, 5, 6, 7
  • 10, 20, 30, 40…
  • 0, 3, 6, 9…
  • 1, 3, 4, 8…

b) Create a finite Arithmetic progression of 5 elements with a=6 and d=3

c) Create an infinite Arithmetic progression with a=7 and d=5

d) Create an arithmetic progression with a=10 and d=(-4)

(use the general Arithmetic Progression formula)

e) Find the common difference and Sn: 8, 16, 24, 32, 40

f) Find the common difference, aₙ and Sn for:

  • 1, 4, 7, 10, 13…(a₈ and S₈)
  • 2, 7, 12, 17…(a₅ and s₆)
  • 3, 4, 5, 6…(a₁₀ and s₉)
  • 2, 4, 6, 8… Find a₈ 

Geometric Progression:

a) Is this a geometric progression? Answer with Yes or No:

  • 2, 6, 18, 54…
  • 1, 3, 5, 7, 9…
  • 2, 4, 8, 12
  • 8, 4, 2, 1

b) Create a Geometric Sequence of 6 elements with a=2 and r=4

c) Create a Geometric Sequence of 6 elements with a=3 and r=(-2)

d) Find the common ratio and Sn: 1, 3, 9, 27, 81

e) Find the common ratio, aₙ and Sn for:

  • 1, 4, 16, 64, 256…(a₈ and S₈)
  • 2, 6, 18, 54…(a₅ and s₆)
  • 3, 9, 27, 81…(a₁₀ and s₉)

Now that you know how the concepts of AP and GP work, try and make more sequences and find the differences, ratios and sums for them. All the best!

Need more practice? Here are some revision notes for 10th Board Exams from India’s some of best ICSE School Teachers. You can order them on Exam18. 

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