The number of terms of a geometric sequence is even.

1. The sum of the odd terms, that is - terms in the odd places (First term + Third Term + Fifth Term + ... and so on) is 4329
2. The sum of the even terms, that is – terms in the even places (Second Term + Fourth Term + Sixth Term + ... and so on) is 5772
3. The last term exceeds the first by 2343

Determine the geometric sequence.

(In reply to Solution by JayDeeKay)

I have solved this puzzle by "guided guessing" as follows: <br>

Since each even-placed member is q times the previous (odd-placed ) member, q = 5771/4329 = 4/3.<br>

So it wasreasonable to assume that the 1st member  is a*3^k, then a*(3^(k-1))*4,  a*(3^(k-2))*4^2, a*(3^(k-2))*4^3,  etc<br>

Looking at tne numbers  given   243 was too small to start with, so I SELECTED a=1,k=6 and got:<p>

 729;  972; 1,296; 1,728; 2,304; 3,072. - the  true results.<br><br>

Truly, nothing warrants  an integer a, but it seemed logical to me

as a 1st choice, and it worked... 

k-1 

Posted by Ady TZIDON on 2014-11-27 13:06:12