SOLUTION: Please help me solve this : In an geometric progression s1=3 , b2=6 Solve b2015

Algebra ->  Expressions-with-variables -> SOLUTION: Please help me solve this : In an geometric progression s1=3 , b2=6 Solve b2015      Log On


   

Question 1012692: Please help me solve this :
In an geometric progression s1=3 , b2=6
Solve b2015
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

In an geometric progression s1=3 , b2=6 
If s1 = 3 then b1 = 3 

r = b2/b1 = 6/3 = 2

sn = b1(rn-1)/(r-1)

s2015 = b1(22015-1)/(2-1)

s2015 = 3(22015-1)/1

s2015 = 3(22015-1) = 11286583986824031675018443500857084975732991938612709298775419160092012655991557133336505414285783235806291210787102354048094558232868179130049475603660667473670634452940778373187709745388589371040949238831200951807011926538294561729477052927882004460755844781653596452514492868751194028396003087676571583906857180428808208934736312874557183646197492269098172683063128145643635487431355396105299678001614677507643449487558749047200550960482207326359475594910311244158234569957999120137592696722223652807184917356582602230498799100175848686444407270658797585246014223236088735742489275966637819607354183778301

Edwin