Sn = a(1-r)^n/(1-r), so
a(1-r)^4/(1-r) ÷ a(1-r)^8/(1-r) = 1/17
(1-r)^-4 = 1/17
(1-r)^4 = 17
1-r = ∜17
r = 1-∜17
I think rather than S4 you meant T4
S4 is usually used for the sum of the first 4 terms.
So, now we have
Tn = ar^(n-1)
T4/T8 = ar^3/ar^7 = r^-4 = 1/17
r^4 = 17
r = ∜17
Still a bit odd. If T4/T8 = 1/16, then we have r=2, and the first three terms are
3,6,12
If I'm off in left field, maybe you can work it from here.
For the geometric series, S4/S8 = 1/17. Determine the first three terms of the series if the first term is 3.
1 answer
1 answer