posted by Crystal .
Determine the sum of the first seven terms of the geometric series in which ...
F) t5 = 5 and t8 = -40
I'm stuck on this one!
t8 = t5 * r^3, so
r^3 = t8/t5 = -40/5 = -8
so, r = -2
t5 = ar^4 = 16a, so a = 5/16
S7 = a(1-r^7)/(1-r)
= 5/16 (1+2^7)/(1+2) = 5/16 * 129/3 = 215/16 = 13.4375
Can you get r by plugging in numbers into an equation, i.e. ar^n-1 and then solving by substitution/elimination?
That was my initial thought but I couldn't figure out how...
Not my orinigal post but the one right above this one, can someone answer?
well, that's basically what I did.
t5 = ar^4
t8 = ar^7
r^3 = t8/t5
a = t5/r^4 = t5 / ∛(t8/t5)^4 = t5 ∛(t5/t8)^4
S7 = t5 ∛(t5/t8)^4 (1-(t8/t5)^(7/3))/(1-∛(t8/t5))
Now just plug in the numbers and let 'er rip!
Thanks again Steve!