in a geometric series t1=23,t3=92 and the sum of all of the terms of the series is 62 813. how many terms are in the series?

Answer 1

for this GS
a = 23
t(3) = ar^2
92 = 23r^2
r^2 = 4
r = ± 2

sum(n) = 62813

sum(n) = a(r^n - 1)/(r-1)
23(2^n -1)/1 = 62813
2^n - 1 = 2731
2^n = 2732

n log2 = log2732
n = 11.4

something wrong here, n has to be a whole number.

check:
for 11 terms,
sum(11) = 23(2^11 - 1)/1 = 47081
for 12 terms
sum(12) = 23(2^12 - 1)/1 = 94185
there is no sum of 62813

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