Asked by Nikhil
Find the values of k for which 6,x,y 162 are geometric progression.
Answers
Answered by
Bosnian
The n-th term of a GP
where:
a = a1 initial value
r = common ratio
an = a r ⁿ ⁻ ¹
In this case a1 = a = 6 , a4 = 162
so
a4 = 6 ∙ r ⁴ ⁻ ¹
162 = 6 ∙162 = 6 ∙ r ³
162 / 6 = r ³
27 = r ³
r = ∛ 27
r = 3
x = a2 = a ∙ r = 6 ∙ 3 = 18
y = a3 = a ∙ r ² = 6 ∙ 3 ² = 6 ∙ 9 = 54
Your GP:
6 , 18 , 54 , 162
where:
a = a1 initial value
r = common ratio
an = a r ⁿ ⁻ ¹
In this case a1 = a = 6 , a4 = 162
so
a4 = 6 ∙ r ⁴ ⁻ ¹
162 = 6 ∙162 = 6 ∙ r ³
162 / 6 = r ³
27 = r ³
r = ∛ 27
r = 3
x = a2 = a ∙ r = 6 ∙ 3 = 18
y = a3 = a ∙ r ² = 6 ∙ 3 ² = 6 ∙ 9 = 54
Your GP:
6 , 18 , 54 , 162
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.