 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Geometric Sequence Settlement III (Posted on 2014-11-26) 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 No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 3
Let a = first term; r = common ration; and 2n = number of terms.

For #1, we have a geometric series with first term = a, common ration = r^2 and number of terms = n.
For #2, we have a geometric series with first term = ar, common ration = r^2 and number of terms = n.

Apply the standard formula for the sum of a geometric series to each of these and divide one by the other: r = 5771/4329 = 4/3.

A bit of algebraic manipulation, after substituting this value for r into each of the 2 equations for the geometric sums, yields n = 3.
Therefore, the original series has 2n = 6 terms.

Substituting these values for r and n into the sum of the odd terms gives a = 729.

So, the original series is: 729; 729*(4/3) = 972; 1,296; 1,728; 2,304; 3,072.

 Posted by JayDeeKay on 2014-11-27 11:41:58 Please log in:

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