Let us carry out an activity. Take a paper and fold it as many times as you can. So how many times did you fold the paper? Maybe four to five times right? Now can you find the height of the stack of the paper after it has been folded several times? Furthermore, how do you determine it? The answer to this is a geometric progression. Let us study this in detail.

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## Geometric Progression Definition

A geometric progression is a sequence in which any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by** r . **For example, the sequence 1, 2, 4, 8, 16, 32… is a geometric sequence with a common ratio of r = 2.

Here the succeeding number in the series is the double of its preceding number. In other words, when 1 is multiplied by 2 it results in 2. When 2 is multiplied by 2 it gives 4. Likewise, when 4 is multiplied by 2 we get 8 and so on. So, what do you think is happening? Can we say that the ratio of the two consecutive terms in the geometric series is constant?

For example, a_{1}= 1, a_{2}= 2

So, \( \frac{a_2}{a_1} \) = \( \frac{2}{1} \) = 2

Also, a_{3 }= 4. Now let us try to find the ratio of a_{3 }and a_{2
}\( \frac{a_3}{a_2} \) = \( \frac{4}{2} \) = 2

Now since the value of a_{4 }= 8

\( \frac{a_4}{a_3} \) = \( \frac{8}{4} \) = 2

So we can say, \( \frac{a_2}{a_1} \) = \( \frac{a_3}{a_2} \) = \( \frac{a_4}{a_3} \)

Thus, the ratio of the two consecutive terms of this particular sequence is a fixed number. Such a sequence is called *Geometric Progression. *Furthermore,* *the geometric progression is the sequence in which the first term is non zero and each consecutive termed is derived by multiplying the preceding term by a fixed quantity.

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**Geometric Progression Example**

Question: Check whether the given sequence, 9, 3, 1, 1/3, 1/9…… is in geometrical progression.

Solution: Let us find the ratio of the consecutive terms, a_{1}= 9 and a_{2 }= 3. So,

\( \frac{a_2}{a_1} \) = \( \frac{3}{9} \) = \( \frac{1}{3} \)

= \( \frac{a_3}{a_2} \) = \( \frac{1}{3} \)

Since the ratio of the consecutive terms of the given sequence is \( \frac{1}{3} \), a fixed number, this sequence is in geometrical progression.

## Geometric Progression Formulas

**The General Form of a Geometric Progression**

The general form of a GP is **a _{1, }a_{1}r, a_{1}r², a_{1}r³, ………, a_{1}r^{n-1}, a_{1 }r^{n}**

**General (n**^{th}) Term of a Geometric Progression

^{th}) Term of a Geometric Progression

If we represent each term of Geometric Progression such as a_{1}, a_{2}, a_{3}, a_{4}….a_{m},….a_{n } according to the first term a_{1}, then the terms in the geometric series will be as follows,

a_{1}= a_{1
}a_{2 }= a_{1}r

a_{3}= a_{2}r = (a_{1}r) r = a_{1}r²

a_{4 }= a_{3}r =( a_{2}r²) = r = a_{1}r³

Likewise, the n^{th }term will be

**a _{n }= a_{1}r^{n-1}**

**Common Ratio of a Geometric Progression**

We get the common ratio of geometric series by dividing any number by its preceding number:

r = \( \frac{a_2}{a_1} \)

**Geometric Progression Sum**

To find the sum of first n term of a GP we use the following formula:

**S _{n }= \( \frac{a_1(1-r^n)}{1-r} \)**

where r ≠1. Here n is the number of terms, a_{1} is the first term and r is the common ratio.

## Properties of Geometric Progression

The following are the properties of G.P:

- If we multiply or divide a non zero quantity to each term of the G.P, then the resulting sequence is also in G.P with the same common difference.
- Reciprocal of all the terms in G.P also form a G.P.
- If all the terms in a G.P are raised to the same power, then the new series is also in G.P.
- If y² = xz, then the three non-zero terms x, y and z are in G.P

## Solved Examples for You

**Q 1: Five terms are in A.P. with common difference ≠0. If the first, third and fourth terms are in G.P then?**

- The fifth term is always zero
- The third term is zero
- The first term is always zero
- Middle term is zero
- Middle term is always 2

**A:** The correct option is A.

**Q 2: How many terms of the series 1 + 3+ 9+…. sum to 121?**

- 5
- 6
- 4
- 3

**A:** The correct option is A. The given series is a G.P,

Sum of n terms of GP is a * (r^{n} – 1)/ (r – 1)

Here a = 1 and r = 3

Substituting values in the equation we get n = 5

**Q 3: Explain what do you understand by geometric progression with example?**

**A:** A geometric progression (GP) is a sequence of terms which differ from each other by a common ratio. For example, the sequence 2, 4, 8, 16 … is a geometric sequence with common ratio 2.

**Q 4: What is the formula to determine the sum in infinite geometric progression?**

**A:** To find the sum of an infinite geometric series that contains ratios with an absolute value less than one, the formula is S=a_{1}/(1−r). Here a_{1} is the first term and r is the common ratio.

**Q 5: Explain the difference between geometric progression and arithmetic progression?**

**A:** A sequence refers to a set of numbers arranged in some specific order. An arithmetic sequence is one where the difference between two consecutive terms is constant. While a geometric sequence is one where the ratio between two consecutive terms is constant.

**Q 6: Can zero be a part of a geometric series?**

**A:** No. If the first term is zero, then geometric progression will not take place.

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