How to Find the Value of n in Geometric Progression ?
In this section, we will learn how to find the value of n in the geometric progression.
Example 1 :
If the geometric sequences
162, 54, 18,......... and 2/81, 2/27, 2/9,..........
have their nth term equal, find the value of n.
Solution :
Let us find nth term of the given sequences one by one.
162, 54, 18,.........
a = 162 r = 54/162 ==> r = 1/3
tn = 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
tn = a r(n-1)
= (2/81) ⋅ (3)(n-1) ------ (2)
tn = tn
162 (3-n+1) = (2/81) x 3n-1
[(162 ⋅ 81)/2] (3-n + 1) = 3n-1
(81 ⋅ 81)(3-n + 1) = 3n-1
(3⁴ x 3⁴)(3-n + 1) = 3n-1
(3⁸)(3-n + 1) = 3n-1
3-n + 9 = 3n-1
-n + 9 = n - 1
2n = 10
n = 5
Therefore the 5th term of the sequence are equal.
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
2r - 5 = 0 2r = 5 r = 5/2 If a = 1 and r = 5/2 a/r = 1/(5/2) a/r = 2/5 a = 1 ar = 1(5/2) ar = 5/2 |
5r - 2 = 0 5r = 2 r = 2/5 If a = 1 and r = 2/5 a/r = 1/(2/5) a/r = 5/2 a = 1 ar = 1(2/5) ar = 2/5 |
Therefore the three terms are
2/5, 1, 5/2 or 5/2, 1, 2/5.
After having gone through the stuff given above, we hope that the students would have understood how to find the value of n in geometric progression.
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