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Geometric Progression 



Problem 2 of 23 


2. For given geometric progression series ,... then find n such that S(n) = .










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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`





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