Posted by **bob** on Monday, June 3, 2013 at 7:54pm.

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

- math -
**Reiny**, Monday, June 3, 2013 at 9:23pm
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 +

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