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- A sequence (t
_{n}) is said to be a geometric progression (G.P.) ift he ratio t_{n + ! }/t_{n}= constant for all n ∈ N - The constant ratio is called the common ratio of the G.P. and is denoted by letter ‘r’. The first term is denoted by the letter ‘a’.

#### General Term or n^{th} Term of a Geometric Progression:

- If ‘a’ is the first term and ‘r’ is the common ratio of the G.P,. then the n
^{th}term is given by

t_{n} = a r^{n- 1}

**Type – I: To Find **t_{n }for Geometric Progression (G.P.).

_{n }for Geometric Progression (G.P.).

#### Example – 01:

- Find n
^{th}term of a geometric progression (G.P.) 2, 6, 18, 54, ….. **Solution:**

Given G.P. is 2, 6, 18, 54, …..

First term = a = 2, common ratio = r = 6/2 = 3

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 2 (3)^{n- 1}

**Ans:** The n^{th} term of the G.P. is 2 (3)^{n- 1}

#### Example – 02:

- Find n
^{th}term of a geometric progression (G.P.) 25, 5, 1, 1/5, ….. **Solution:**

Given G.P. is 25, 5, 1, 1/5, …..

First term = a = 25, common ratio = r = 5/25 = 1/5

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 25 (1/5)^{n- 1 }= 5^{2} (1/5)^{n- 1 } = (1/5)^{-2} (1/5)^{n- 1} = (1/5)^{n- 3}

**Ans:** The n^{th} term of the G.P. is (1/5)^{n- 3}

#### Example – 03:

- Find n
^{th}term of a geometric progression (G.P.) 3, 3, 3, 3, ….. **Solution:**

Given G.P. is 3, 3, 3, 3, …..

First term = a = 3, common ratio = r = 3/3 = 1

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 3 (1)^{n- 1 }= 3 x 1 = 3

**Ans:** The n^{th} term of the G.P. is 3

#### Example – 04:

- Find n
^{th}term of a geometric progression (G.P.) 3, 15, 75, 375, ….. **Solution:**

Given G.P. is 3, 15, 75, 375, …..

First term = a = 3, common ratio = r = 15/3 = 5

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 3(5)^{n- 1}

**Ans:** The n^{th} term of the G.P. is 3(5)^{n- 1}

#### Example – 05:

- Find n
^{th}term of a geometric progression (G.P.) 1, -4, 16, -64, ….. **Solution:**

Given G.P. is 1, -4, 16, -64, …..

First term = a = 1, common ratio = r = -4/1 = – 4

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 1(- 4)^{n- 1 }= (- 4)^{n- 1}

**Ans:** The n^{th} term of the G.P. is (- 4)^{n- 1}

#### Example – 06:

- Find n
^{th}term of a geometric progression (G.P.) 1, -3/2, 9/4, – 27/8, ….. **Solution:**

Given G.P. is 1, -3/2, 9/4, – 27/8, …..

First term = a = 1, common ratio = r = (-3/2)/1 = – 3/2

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 1(- 3/2)^{n- 1 }= (- 3/2)^{n- 1}

**Ans:** The n^{th} term of the G.P. is (- 3/2)^{n- 1}

#### Example – 06:

- Find n
^{th}term of a geometric progression (G.P.) √3, 1/√3, 1/3√3, 1/9√3, ….. **Solution:**

Given G.P. is √3, 1/√3, 1/3√3, 1/9√3, …..

First term = a = √3, common ratio = r = (1/√3)/√3= 1/3

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= √3(1/3)^{n- 1 }= (3)^{1/2}(1/3)^{n- 1} = (1/3)^{-1/2}(1/3)^{n- 1} = (1/3)^{n- 3/2}

**Ans:** The n^{th} term of the G.P. is (1/3)^{n- 3/2}

#### Example – 07:

- Show that the sequence 3, 6, 12, 24, 48, ….. is a G.P. Find the 7
^{th}term. **Solution:**

Given sequence is 3, 6, 12, 24, 48, …..

t_{2}/t_{1} = 6/3 = 2, t_{3}/t_{2} = 12/6 = 2, t_{4}/t_{3} = 24/12 = 2, t_{5}/t_{14} = 48/24 = 2,

Thus in given sequence the ratio of next term to the previous term is constant and is equal to 2

Hence the given sequence is a G.P.

First term = a = 3, common ratio = r = 2

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 3(2)^{n- 1}

∴ t_{7} = a r^{n- 1 }= 3(2)^{7- 1 }= 3(2)^{6 }= 3 x 64 = 192

**Ans:** The 7^{th} term of the G.P. is 192

#### Example – 08:

- If a = 7, r = 1/3, find t
_{6}. **Solution:**

Given first term = a = 7, common ratio = r = 1/3

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 7(1/3)^{n- 1}

∴ t_{6} = 7(1/3)^{6- 1}^{ }= 7(1/3)^{5 }= 7(1/243)^{ }= 7/243

**Ans:** t_{6}. = 7/243

#### Example – 09:

- If a = 5, r = – 2, find t
_{5}. **Solution:**

Given first term = a = 5, common ratio = r = – 2

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{n} = a r^{n- 1 }= 5(- 2)^{n- 1}

∴ t_{5} = 5(- 2)^{5- 1}^{ }= 5(-2)^{4 }= 5 x 16 = 80

**Ans: **t_{5}. = 80

#### Example – 10:

- If a = 2/3, t
_{6}= 162, find r. **Solution:**

Given first term = a = 2/3, t_{6} = 162

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{6} = (2/3) r^{6- 1 }

∴ 162 = (2/3) r^{5 }

∴ (162 x 3)/2= r^{5}

∴ r^{5 }= 81 x 3= 243^{ }

∴ r = 3

**Ans:** r = 3

#### Example – 11:

- If r = 2, t
_{8}= 640, find a. **Solution:**

Given fcommon ratio = r = 2, t_{8} = 640

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{8} = a (2)^{8- 1 }

∴ 640 = a x 2^{7 }

∴ 640 = 128 a

∴ a = 5

**Ans:** a = 5

#### Example – 12:

- If a = 5, t
_{6}= 1/625, find r and t_{10}. **Solution:**

Given first term = a = 5, t_{6} = 1/625

the n^{th} term of a G.P. is given by t_{n} = a r^{n- 1}

∴ t_{6} = 5 r^{6- 1 }

∴ 1/625 = (5) r^{5 }

∴ 1/3125= r^{5}

∴ r^{ }= 1/5^{ }

Now, t_{n} = a r^{n- 1}

t_{10} = 5(1/5)^{10 – 1 } (1/5)^{– 1}(1/5)^{9 }= (1/5)^{8} = 1/390625

**Ans:** r = 1/5 and t_{10} = (1/5)^{8} = 1/390625

#### Example – 13:

- If for a sequence, t
_{n}= 4^{n – 3}/5^{n – 2}, show that the sequence is G.P. and find the first term and common ratio. **Solution:**

Thus in given sequence the ratio of next term to the previous term is constant and is equal to 4/5

Hence the given sequence is a G.P. and its common ratio = r = 4/5

the n^{th} term of a G.P. is given as t_{n} = 4^{n – 3}/5^{n – 2}

t_{1} = 4^{1 – 3}/5^{1 – 2 }= 4^{-2}/5^{-1} = 5^{1}/4^{2} = 5/16

**Ans:** First term = 5/16 and common ratio = 4/5

#### Example – 14:

- If for a sequence, t
_{n}= 2^{n – 2}/5^{n – 3}, show that the sequence is G.P. and find the first term and common ratio. **Solution:**

Thus in given sequence the ratio of next term to the previous term is constant and is equal to 2/5

Hence the given sequence is a G.P. and its common ratio = r = 25

the n^{th} term of a G.P. is given as t_{n} = 2^{n – 2}/5^{n – 3}

t_{1} = 2^{1 – 2}/5^{1 – 3}^{ }= 2^{-1}/5^{-2} = 5^{2}/2^{1} = 25/2

**Ans:** First term = 25/2 and common ratio = 2/5

#### Example – 15:

- If for a sequence, t
_{n}= (-5)^{n +1}/3^{n – 1}, show that the sequence is G.P. and find the first term and common ratio. **Solution:**

Thus in given sequence the ratio of next term to the previous term is constant and is equal to – 5/3

Hence the given sequence is a G.P. and its common ratio = r = – 5/3

the n^{th} term of a G.P. is given as t_{n} = (-5)^{n +1}/3^{n – 1}

t_{1} = (-5)^{1 +1}/3^{1 – 1}^{ }= (-5)^{2}/3^{0} = 25/1 = 25

**Ans:** First term = 25 and common ratio = – 5/3

Science > Mathematics > Sequence and Series > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |

If a quadratic equation…

ax²+bx+c=0 has equal roots then prove that a, b/2& c are in G. P….

We shall cover such problems in coming articles. Thank you for your suggestion