Consider the following sequences.

a. 2, 4, 8, 16, … … … …

b. 27, 9, 3, 1, … … … …

c. 16, -24, 36, -54, … … … …

In the first sequence, each term is double of its previous term.
In the second sequence, each term is one-third of its previous term. Each term
of the third sequence is -3/2 times its previous term. In each of the above
sequence, the ratio of a term and its previous term is equal or constant. Such
a sequence is known as a **Geometric Sequence** or **Geometric Progression**.

**Definition**

A sequence is said to be a **geometric
sequence** or **geometric progression**
if the ratio between any term and its previous term is equal or constant
throughout the whole sequence. It is denoted by GP.

The constant ratio obtained by dividing a term by its previous
term is called common ratio of the sequence. In above example 2, 4, 8, 16, … … the
common ratio is 4/2 = 8/4 = 16/8 = 2. Similarly, the common ratio of 27, 9, 3,
1, … … is 1/3 and that of 16, -24, 36, -54, … …
is -3/2. The common ratio of a geometric sequence is denoted by r.

A series corresponding to any geometric sequence is known as the
**geometric series** associated with the
given geometric sequence. Hence, 2 + 4 + 8 + 16 + … … … is a geometric series
associated with the geometric sequence 2, 4, 8, 16, … … … .

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**General Term of a GP**

If a be the first term and r be the common ratio of a geometric
progression, then the terms of the progression are: a, ar, ar^{2}, ar^{3},
… … …

If t_{1}, t_{2}, t_{3}, t_{4}, …
… …, t_{n} be the first, second, third, fourth, … … …, n^{th} terms of a
GP respectively then,

t_{1} = a = ar^{1-1}

t_{2} = ar = ar^{2-1}

t_{3} = ar^{2} = ar^{3-1}

t_{4} = ar^{3} = ar^{4-1}

… … … … … … …

t_{n} = ar^{n-1}

Thus, if the first term and the common ratio of a GP are known,
we can find any term.

**Worked Out Examples**

**Example1:** If the first term and the common ratio of a GP are 1/8 and -2
respectively, find the seventh term of the GP.

**Solution:**

Here,

First term (a) = 1/8

Common ratio (r) = -2

Seventh term (t_{7}) = ?

By using formula, we have,

t_{n} = ar^{n-1}

or, t_{7} = 1/8 × (-2)^{7-1}

or, t_{7} = 1/8 × (-2)^{6}

or, t_{7} = 1/8 × 64

or, t_{7} = 8

∴ Seventh term t_{7} = 8.

**Example 2:** For a geometric series 1/18 + 1/6 + 1/2 + … … … + 81/2

a. Find the number of terms.

b. Find the 9^{th} term of the series.

**Solution:**

a.

Here,

The first term (a) = 1/18

Common ratio (r) = (1/6)/(1/18) = 3

Last term (t_{n}) = 81/2

Number of terms (n) = ?

By using formula,

t_{n} =ar^{n-1}

or, 81/2 = 1/18 × 3^{n-1}

or, 81 × 9 = 3^{n-1}

or, 3^{n-1} = 3^{6}

or, n – 1 = 6

or, n = 7

∴ Number of terms n = 7

b.

9^{th} term (t_{9}) = ?

By using formula,

t_{n} = ar^{n-1}

or, t_{9} = 1/18 × 3^{9-1}

or, t_{9} = 1/18 × 3^{8}

or, t_{9} = 1/18 × 6561

or, t_{9} = 729/2

∴ 9^{th} term t_{9} = 729/2

**Example 3:** Is 1458 a term of the series 2 + 6 + 18 + 54 + … … … ?

**Solution:**

Here,

The series is 2 + 6 + 18 + 54 + … … …

The first term (a) = 2

Common ratio (r) = 6/2 = 3

Suppose, t_{n} = 1458

or, ar^{n-1} = 1458

or, 2 × 3^{n-1} = 1458

or, 3^{n-1} = 1458/2

or, 3^{n-1} = 729

or, 3^{n-1} = 3^{6}

or, n – 1 = 6

or, n = 7

∴ 1458 is the 7^{th} term of the given series.

**Example 4:** Find the value of x for which x + 9, x – 6, 4 are the three terms
of a geometric progression and calculate the fourth term of geometric
progression in each case.

**Solution:**

The given three terms in GP are x + 9, x – 6, 4.

From the definition of GP,

^{2}= 4(x + 9)

or, x^{2} – 12x + 36 = 4x + 36

or, x^{2} – 16x = 0

or, x(x – 16) = 0

∴ x = 0 or 16

__Case I__,

When x = 0, the given numbers are 9 , -6, 4

First term (a) = 9

Common ratio (r) = -2/3

Fourth term (t_{4}) = ?

We have, t_{n} = ar^{n-1}

or, t_{4} = 9 × (-2/3)^{4-1}

or, t_{4} = 9 × (-2/3)^{3}

or, t_{4} = 9 × -8/27

∴ t_{4} = -8/3

__Case II__,

When x = 16, the given numbers are 25, 10, 4

First term (a) = 25

Common ratio (r) = 2/5

Fourth term (t_{4}) = ?

We have, t_{n} = ar^{n-1}

or, t_{4} = 25 × (2/5)^{4-1}

or, t_{4} = 25 × (2/5)^{3}

or, t_{4} = 25 × (8/125)

∴ t_{4} = 8/5

**Example 5:** If the fourth and tenth terms are 32 and 1/2 respectively, find the
GP. Also find its 15^{th} term.

**Solution:**

Let a be the first term and r be the common ratio of the GP.

Fourth term (t_{4}) = 32

Tenth term (t_{10}) = ½

We have,

t_{n} = ar^{n-1}

or, t_{4} = ar^{4-1}

or, ar^{3} = 32 … … … (i)

and, t_{10} = ar^{10-1}

or, ar^{9} = 1/2 … … … (ii)

Dividing equation (ii) by (i),

__Case I__,

When r = ½

ar^{3} = 32

or, a × 1/8 = 32

or, a = 256

∴ The given GP is 256, 128, 64, … … …

And,

t_{15} = ar^{15-1}

or, t_{15} = 256 × (1/2)^{14}

or, t_{15} = 256 × 1/16384

∴ t_{15} = 1/64

__Case II__,

When r = -1/2

ar_{3} = 32

or, a × (-1/2)^{3} = 32

or, a × -1/8 = 32

or, a = -256

The given GP is -256, 128, -64, … … …

And,

t_{15} = ar^{15-1}

or, t_{15} = -256 × (-1/2)^{14}

or, t_{15} = -256 × 1/16384

∴ t_{15} = -1/64

**Example 6:** In a geometric series, 7^{th} term is 16 times of the third
term and the fifth term is 1/16, find the second term.

**Solution:**

Here,

t_{7} = 16 × t_{3}

or, ar^{6} = 16 × ar^{2}

or, ar^{6}/ar^{2} = 16

or, r^{4} = 16 = 2^{4}

∴ r = 2.

Now,

t_{5} = 1/16

or, ar^{4} = 1/16

or, ar × r^{3} = 1/16

or, t_{2} × 2^{3} = 1/16

or, t_{2} × 8 = 1/16

or, t_{2} = 1/128

∴ Second term t_{2} = 1/128

**Example 7:** The product of three numbers in GP is 1000. If we add 6 to its
second number and 7 to its third number, the resulting three numbers form an AP.
Find the numbers in GP.

**Solution:**

Let the numbers in GP be a/r, a and ar.

Then, a/r × a × ar = 1000

or, a^{3} = 1000

∴ a = 10

Hence the numbers are 10/r, 10 and 10r.

The new numbers are 10/r, 10 + 6, and 10r + 7 and these numbers
are in AP.

∴ 16 – 10/r = (10r + 7) – 16

or, 16r – 10 = 10r^{2} – 9r

or, 10r^{2} – 25r + 10 = 0

or, 2r^{2} – 5r + 2 = 0

or, 2r^{2} – 4r – r + 2 = 0

or, 2r(r – 2) – 1(r – 2) = 0

or, (r – 2)(2r – 1) = 0

∴ r = 2 or 1/2

__Case I__,

When r = 2, the numbers are 10/2, 10, 10×2 i.e. 5, 10, 20

__Case II__,

When r = 1/2, the numbers are 10/(1/2), 10, 10×1/2 i.e. 20, 10,
5

Hence, the numbers are 5, 10, 20 or 20, 10, 5.

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