math
posted by bob .
in a geometric series t1=23,t3=92 and the sum of all of the terms of the series is 6813. How many terms are in the series

t1 = 23 > a = 23
t3 = 92 >ar^2 = 92
divide them
r^2 = 4
r = ± 2
sum(n) = a(r^n  1)/(r1) = 6813
If r = 2
23( 2^n  1)/(21) = 6813
2^n  1 = 296.21... > not a whole number
so there can't be a sum of 6813
if r = 2
23((2)^n  1)/(21) = 6813
(2)^n = 887.65...
no way!
This question either has a typo , or the question itself is flawed.
proof:
suppose we had 4 terms
sum(4) = 23(2^4  1)/1 = 345
sum(5) = 23(2^5  1)/1 = 713
...
sum(8) = 5865 
sum(9) = 11753
or
sequence is
23 + 46 + 92 + 184 + 368 + 736 + 1472 + 2944 + 5888 +