# General Term of Geometric Progression (G.P.)

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• A sequence (tn) is said to be a geometric progression (G.P.) ift he ratio tn + ! /tn = 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 nth Term of a Geometric Progression:

• If ‘a’ is the first term and ‘r’ is the common ratio of the G.P,. then the nth term is given by

tn = a rn- 1

### Type – I: To Find tn for Geometric Progression (G.P.).

#### Example – 01:

• Find nth 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 nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 2 (3)n- 1

Ans: The nth term of the G.P. is 2 (3)n- 1

#### Example – 02:

• Find nth 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 nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 25 (1/5)n- 1  = 52 (1/5)n- 1  = (1/5)-2 (1/5)n- 1 = (1/5)n- 3

Ans: The nth term of the G.P. is (1/5)n- 3

#### Example – 03:

• Find nth 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 nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 3 (1)n- 1 = 3 x 1 = 3

Ans: The nth term of the G.P. is 3

#### Example – 04:

• Find nth 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 nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 3(5)n- 1

Ans: The nth term of the G.P. is 3(5)n- 1

#### Example – 05:

• Find nth 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 nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 1(- 4)n- 1 = (- 4)n- 1

Ans: The nth term of the G.P. is (- 4)n- 1

#### Example – 06:

• Find nth 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 nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 1(- 3/2)n- 1 = (- 3/2)n- 1

Ans: The nth term of the G.P. is (- 3/2)n- 1

#### Example – 06:

• Find nth term of a geometric progression (G.P.) 3, 1/3, 1/33, 1/93,  …..
• Solution:

Given G.P. is 3, 1/3, 1/33, 1/93,  …..

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

the nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 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 nth 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 7th term.
• Solution:

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

t2/t1 = 6/3 = 2,  t3/t2 = 12/6 = 2,  t4/t3 = 24/12 = 2,  t5/t14 = 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 nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 3(2)n- 1

∴ t7 = a rn- 1  = 3(2)7- 1 = 3(2)= 3 x 64 = 192

Ans: The 7th term of the G.P. is 192

#### Example – 08:

• If a = 7, r = 1/3, find  t6.
• Solution:

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

the nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 7(1/3)n- 1

∴ t6 = 7(1/3)6- 1  = 7(1/3)= 7(1/243) = 7/243

Ans: t6. = 7/243

#### Example – 09:

• If a = 5, r = – 2, find  t5.
• Solution:

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

the nth term of a G.P. is given by tn = a rn- 1

∴ tn = a rn- 1  = 5(- 2)n- 1

∴ t5 = 5(- 2)5- 1  = 5(-2)= 5 x 16 = 80

Ans: t5. = 80

#### Example – 10:

• If a = 2/3, t6 = 162, find r.
• Solution:

Given first term = a = 2/3, t6 = 162

the nth term of a G.P. is given by tn = a rn- 1

∴ t6 = (2/3) r6- 1

∴ 162 = (2/3) r5

∴ (162 x 3)/2=  r5

∴ r= 81 x 3=  243

∴ r = 3

Ans: r = 3

#### Example – 11:

• If r = 2, t8 = 640, find a.
• Solution:

Given fcommon ratio = r = 2, t8 = 640

the nth term of a G.P. is given by tn = a rn- 1

∴ t8 = a (2)8- 1

∴ 640 = a x 27

∴ 640 = 128 a

∴ a = 5

Ans: a = 5

#### Example – 12:

• If a = 5, t6 = 1/625, find r and t10.
• Solution:

Given first term = a = 5, t6 = 1/625

the nth term of a G.P. is given by tn = a rn- 1

∴ t6 = 5 r6- 1

∴ 1/625 = (5) r5

∴ 1/3125=  r5

∴ r = 1/5

Now, tn = a rn- 1

t10 = 5(1/5)10 – 1  (1/5)– 1(1/5)= (1/5)8 = 1/390625

Ans: r = 1/5 and t10 = (1/5)8 = 1/390625

#### Example – 13:

• If for a sequence, tn = 4n – 3/5n – 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 nth term of a G.P. is given as tn = 4n – 3/5n – 2

t1 = 41 – 3/51 – 2 = 4-2/5-1  =  51/42 = 5/16

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

#### Example – 14:

• If for a sequence, tn = 2n – 2/5n – 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 nth term of a G.P. is given as tn = 2n – 2/5n – 3

t1 = 21 – 2/51 – 3 = 2-1/5-2  =  52/21 = 25/2

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

#### Example – 15:

• If for a sequence, tn = (-5)n +1/3n – 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 nth term of a G.P. is given as tn = (-5)n +1/3n – 1

t1 = (-5)1 +1/31 – 1 = (-5)2/30  = 25/1 = 25

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

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