FIND GEOMETRIC SEQUENCE FROM THE GIVEN TWO TERMS

Example 1 :

If the 4^th and 7^th terms of a G.P are 54 and 1458 respectively, find the G.P

Solution :

4^th term  =  54

7^th term  =  1458

t₄ = 54

a r³  =  54   ----- (1)

t₇  =  1458

a r⁶  =  1458  ----- (2)

(2)/(1)  =  (a r⁶)/(a r³)  ==>  1458/54

r³  =  27

r  =  3

By substituting r  =  3 in the first equation we get 

a (3)³  =  54

a(27)  =  54 

a  =  54/27

 a  =  2  

The general form of G.P is   a, a r , a r ²,.........

  =  2, 2(3), 2(3)²,..............

  =  2,6,18,............

Therefore the required geometric sequence is

2, 6, 18, .......

Example 2 :

In a geometric sequence, the first term is 1/3 and the sixth term is 1/729, find the G.P

Solution :

1^st term  =  1/3

6^th term = 1/729

a  =  1/3

t₆  =  1/729

a r⁵  =  1/729  ------(1)

Applying the value of a in (1) we get, 

(1/3) r⁵  =  1/729

r⁵  =  (1/729) / (1/3)

r⁵  =  (1/729) ⋅ (3/1)

r⁵  =  1/243

r⁵  =  (1/3)⁵

r  =  1/3

The general form of G.P is   a, ar, ar²,.........

  =  1/3, (1/3) (1/3), (1/3)(1/3)²,..............

  =  1/3, 1/9, 1/27,............

Example 3 :

The fifth term of a G.P is 1875.If the first term is 3,find the common ratio.

Solution :

Fifth term (t₅)  =  1875

ar⁴  =  1875 ----(1)

First term (a)  =  3

By applying the value of a in (1), we get 

3r⁴  =  1875

r⁴  =  1875/3

r⁴  =  625

r⁴  =  5⁴

r  =  5

Therefore the common ratio is 5.

Example 4 :

Find the G.P where the 4^th term is 8 and 8^th term is 128/625.

Solution :

4^th term  = 8

8^th term = 128/625

t₄ = 8

a r³  =  8   ----- (1)

t₈  =  128/625

a r⁷  =  128/625 ----- (2)

(2)/(1) ==> a r⁷/a r³  =  (128/625) / (8/1)

= (128/625) x (1/8)

Doing simplification, we get

r⁴ = 16/625

r⁴ = (2/5)⁴ 

r = 2/5

Applying the value of r in (1), we get

a (2/5)³  =  8

a = 8(125/8)

a = 125

ar = 125 (2/5) ==> 50

ar² = 50 (2/5) ==> 20

So, the required G.P is 125, 50, 20,..........

Example 5 :

Insert 3 geometric mean between 1/9 and 9.

Solution :

Since we have 3 terms in between 1/9 and 9.

The first term of the sequence (a) = 1/9

second term = ar

third term = ar²

fourth term = ar³

fifth term = 9 = ar⁴

ar⁴ = 9

Applying the value of a, we get

(1/9)r⁴ = 9

r⁴ = 9

r = 3

Applying the values of a and r in the following,

second term = ar  ==> (1/9) x 3 ==> 1/3

third term = ar² ==> (1/3) x 3 ==> 1

fourth term = ar³ ==> 1 x 3 ==> 3

The required geometric means between 1/9 and 9 are 1/3, 1, 3

Example 6 :

Find three numbers in G.P whose sum is 19 and product is 216.

Solution :

Let a/r, a and ar be three terms of G.P

Sum of the terms = 19

Product of the terms = 216

(a/r) + a + ar = 19

(a/r) x a x ar = 216

a³ = 216

a = 6

6r² - 13r + 6 = 0

6r² - 9r - 4r + 6 = 0

3r(2r - 3) - 2(2r - 3) = 0

(3r - 2) (2r - 3) = 0

r = 2/3 and r = 3/2

Applying a = 6 and r = 2/3, we get

a/r = 6 / (2/3) ==> 6 x (3/2) ==> 9

a = 6

ar = 6 (2/3) ==> 4

So, the three terms are 9, 6, 4 (or) 4, 6, 9

Example 7 :

The second term of G.P is 24 and fifth term is 81. The series is___?

Solution :

Second term = 24

Fifth term = 81

ar = 24 ----(1)

ar⁴ = 81 ----(2)

ar (r³) = 81

24(r³) = 81

r³ = 81 / 24

r³ = 27/8

r = 3/2

Applying the value of r in (1), we get

a(3/2) = 24

a = 24(2/3)

a = 16

To find the required G.P, let us find the first three terms of Geometric progression,

ar = 16(3/2) ==> 24

ar² = 24(3/2) ==> 36

So, the required G.P is 16, 24, 36, .............

Example 8 :

In a geometric progression the product of first three terms is 27/8. The middle terms is 

Solution :

Let the first three terms of the geometric progression be 

a/r, a and ar

Product of these three terms = 27/8

(a/r) x a x ar = 27/8

a³ = 27/8

a³ = (3/2)³

a = 3/2

So, the required middle term is 3/2.

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