|
|
|
|
|
|
|
|
|
|
Method and examples
|
|
|
|
Geometric Progression |
|
|
|
|
|
|
Problem 3 of 23 |
|
|
|
3. For given geometric progression series ,... then find n such that f(n) = .
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Solution
|
Solution provided by AtoZmath.com
|
|
|
This is demo example. Please click on Find button and solution will be displayed in Solution tab (step by step)
| Geometric Progression |
3. For given geometric progression series 3,6,12,24,48 ,... then find n such that f(n) = 1536 .
Here `a = 3,`
`r = 6/3 = 2`
Let n be the term such that `f(n) = 1536`
We know that, `a_n = a × r^(n-1)`
`=> 3 × 2^(n-1) = 1536`
`=> 2^(n-1) = 512`
`=> 2^(n-1) = 2^9`
`=> n - 1 = 9`
`=> n = 9 + 1`
`=> n = 10`
|
|
|
|
|
|
|
Share this solution or page with your friends.
|
|
|
|
