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

Answer 1

t1 = 23 ---> a = 23
t3 = 92 --->ar^2 = 92
divide them
r^2 = 4
r = ± 2

sum(n) = a(r^n - 1)/(r-1) = 6813

If r = 2
23( 2^n - 1)/(2-1) = 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)/(-2-1) = 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 +