HOW TO FIND THE VALUE OF N IN GEOMETRIC PROGRESSION

Example 1 :

If the geometric sequences

162, 54, 18,......... and 2/81, 2/27, 2/9,..........

have their n^th term equal, find the value of n.

Solution :

Let us find n^th term of the given sequences one by one.

162, 54, 18,.........

a  =  162   r  =  54/162 ==> r = 1/3 

t_n  =  a r^(n-1)

  = 1 62 (1/3)^(n-1)

  =  162 (3^-1)^(n-1)

  =  162 (3^-n+1)  ------ (1) 

2/81, 2/27, 2/9,..........

a  =  2/81, r  =  (2/27) / (2/81)

r  =  (2/27) ⋅ (81/2)

r = 3

t_n = a r^(n-1)

  =  (2/81) ⋅ (3)^(n-1)  ------ (2)

t_n  =  t_n 

162 (3^-n+1)  =  (2/81) x 3^n-1

[(162 ⋅ 81)/2] (3^{-n + 1})  =  3^n-1

(81 ⋅ 81)(3^{-n + 1})  =  3^n-1

(3⁴ x 3⁴)(3^{-n + 1})  =  3^n-1

(3⁸)(3^{-n + 1})  =  3^n-1

3^{-n + 9}  =  3^n-1

-n + 9  =  n - 1

2n  =  10

n  =  5

Therefore the 5th term of the sequence are equal.

Find the First Three Terms of a Geometric Sequence When Sum of Three Terms is Given

Example 2 :

The sum of three terms of a geometric sequence is 39/10 and their product is 1. Find the common ratio and the terms.

Solution :

Let the first three terms are a/r, a, ar

Sum of three terms  =  39/10

Product of three terms  =  1

(a/r) + a + a r  =  39/10   ------ (1)

(a/r) ⋅ a ⋅ a r = 1

a³  =  1³

 a  =  1

By applying the value of a in the first equation, we get

(1/r) + 1 + 1 r  =  39/10

 (1 + r + r²)/r  =  39/10

10(1 + r + r²)  =  39 r

10r² + 10r - 39r + 10  =  0

10 r² - 29 r + 10  =  0

(2r - 5)(5r - 2)  =  0

Therefore the three terms are

2/5, 1, 5/2 or 5/2, 1, 2/5.

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