Question

Question #572cf

Answer 1

`sum_(n=0)^(49)10*(0.8)^n ~~ 49.99928638`

Explanation:

Sum of Infinite series is `S_oo= a/(1-r); |r| <1` where `a`

is first term and `r` is the common ratio. `a=10; S_oo=50`

`S_oo= a/(1-r) or 50 = 10/(1-r) or 50-50r =10` or

`50r=40 :. r=40/50=0.8`

Geometric series is `10 , (10*0.8), (10*0.8^2),(10*0.8^3)...`

`sum_(n=0)^(49)10*(0.8)^n =a *(1-r^n)/(1-r)` or

`sum_(n=0)^(49)10*(0.8)^n =10 *(1-(0.8)^n)/(1-0.8)` or

`sum_(n=0)^(49)10*(0.8)^n ~~ 49.99928638` [Ans]