Question

Question #c25ad

Answer 1

`Sigma_oo = 80`

Explanation:

find sums in terms of `a` and `r`:
`u_n = ar^(n-1)`

`u_1+u_3 = ar^0 + ar^2 = 150`

`r^0 = 1 -> ar^0 = a`
`ar^0 + ar^2 = a + ar^2 = a(1+r^2)`

`u_2 + u_4 = -75`

`ar^1 + ar^3 = ar + ar^3 = ar(1+r^2)`

solve, using values given, to find `r` and `a:`

`a(1+r^2) = 150`
`ar(1+r^2) = -75`

`r = (ar(1+r^2))/(a(1+r^2))`

`r = -75/150 = -1/2`

`a(1+r^2) = 150`

`a(1+(-1/2)^2) = 150`

`a (1+1/4) = 150`

`a* 5/4 = 150`

`a = 4/5 * 150 = 120`

`r = -1/2, a = 120`

substitute `r` and `a` into the formula for infinite geometric series:

`Sigma_oo = a/(1-r)`

`a/(1-r) = 120/(1-(-1/2))`

`120/1.5 = 240/3 = 80`

`Sigma_oo = 80`, when `u_n = 120 * (-1/2)^(n-1)`