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Geometric Sequence Settlement III (Posted on 2014-11-26) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
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Solution A slightly different approach Comment 3 of 3 |
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^(n-1) = a + 2343
So r^(n-1) = 1 + 2343/a
or r^n = 4/3 + 3124/a

The sum of the series can be given by the formula a(1-r^n)/(1-r)
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 2014-11-28 16:13:21
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