|
|
|
|
|
|
|
|
|
Method and examples
|
|
|
|
Geometric Progression |
|
|
|
|
|
|
Problem 2 of 23 |
|
|
|
2. For given geometric progression series ,... then find n such that S(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 |
2. For given geometric progression series 3,6,12,24,48 ,... then find n such that S(n) = 3069 .
Here `a = 3,`
`r = 6/3 = 2`
We know that, `S_n = a * (r^n - 1)/(r - 1)`
`=> S_n = 3 × ((2)^n - 1) / (2 - 1)`
`=> 3069 = 3 × ((2)^n - 1)/(1)`
`=> 2^n - 1 = 3069 × (1) / 3`
`=> 2^n - 1 = 1023`
`=> 2^n = 1024`
`=> 2^n = 2^10`
`=> n = 10`
|
|
|
|
|
|
|
Share this solution or page with your friends.
|
|
|
|
