The number of terms of a geometric sequence is even.
 The sum of the odd terms, that is  terms in the odd places (First term + Third Term + Fifth Term + ... and so on) is 4329
 The sum of the even terms, that is – terms in the even places (Second Term + Fourth Term + Sixth Term + ... and so on) is 5772
 The last term exceeds the first by 2343
Determine the geometric sequence.
Clearly dividing the terms gives the ratio r = 5772/4329 = 4/3
Summing all terms gives 4329+5772 = 10101
Call the first term a
the last term is a*r^(n1) = a + 2343
So r^(n1) = 1 + 2343/a
or r^n = 4/3 + 3124/a
The sum of the series can be given by the formula a(1r^n)/(1r)
subbing the above gives
a(1  (4/3 + 3124/a)/(1(4/3)) = 10101
a(1/3  3124/a) = 3367
a/3  3124 = 3367
a/3 = 243
a = 729
To find n use r^n = 4/3 + 3124/a
(4/3)^n = 4/3 + 3124/729 = 4096/729 = 4^6/3^6 = (4/3)^6
so n=6
The series is then
729, 972, 1296, 1728, 2304, 3072

Posted by Jer
on 20141128 16:13:21 