PROBLEMS ON GEOMETRIC SEQUENCE

Problem 1 :

In a geometric sequence. the 9^th term is 32805 and 6^th term is 1215. Find the 12^th term.

Solution :

(2) / (1)  ==> ar⁵/ar⁸  =  1215/32805

1/r³  =  1/27

(1/r)³  =  (1/3)³

r = 3

By applying the value of r in (2), we get

a(3)⁵  =  1215  

a  =  1215/3⁵

a  =  5(3⁵)/3⁵

a = 5

12th term :

t₁₂  =  ar¹¹

t₁₂   =  5(3)¹¹

Problem 2 :

Find the 10^th term of a geometric sequence whose 8^th term is 768 and the common ratio is 2.

Solution :

8^th term is 768

t₈  =  768

ar⁷  =  768

r = 2

a(2⁷)  =  768

a  =  768/128

a = 6

10th term :

t₁₀  =  ar⁹

=  6(2⁹)

  =  6(512)

t₁₀  =  3072

Hence the 10^th term of the sequence is 3072.

Problem 3 :

If a, b, c are in arithmetic sequence, then show that 3^a, 3^b, 3^c are in geometric sequence.

Solution :

If a, b, c are in arithmetic sequence, then b = (a + c)/2

In geometric sequence,  b  =  √ac

To prove that 3^a, 3^b, 3^c are in G.P.

3^b  =  √(3^a ⋅ 3^c) 

3^b  =  (3^{a + c})^1/2

3^b  =  3^(a+c)/2

b = (a + c)/2

Hence 3^a, 3^b, 3^c are in geometric sequence.

Problem 4 :

In a geometric sequence, the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 57/2 . Find the three terms.

Solution :

Let the three terms be a/r, a, ar

The product of three consecutive terms  =  27

(a/r) ⋅ a ⋅ ar  =  27

a³  =  27

a  =  3

Sum of the product of two terms taken at a time  =  57/2

[(a/r) ⋅ a] + [a ⋅ ar] + [ar ⋅ a/r]  =  57/2

a²/r + a²r + a²  =  57/2

a²(1/r + r + 1)  =  57/2

9(1 + r + r²)/r  =  57/2

18(r² + r + 1)  =  57 r

18r² + 18r + 18  =  57 r

18r² + 18r - 57r + 18  =  0

18r²  - 39r + 18  =  0

6r²  - 13r + 6  =  0

(2r - 3)(3r - 2)  =  0

r = 3/2 and r = 2/3

First term  =  a/r  =  3/(3/2)  =  2

Second term  =  a  =  3  =  3

Third term  =  ar  =  3(3/2)  =  9/2

Hence the required three terms are 2, 3, 9/2 (or) 9/2, 3, 2.

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