 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Geometric and Three Power Travail (Posted on 2016-02-22) S(0), S(1), S(2) , ..... are terms of a geometric sequence in strictly ascending orders of magnitude.
All the terms of this sequence are nonnegative integer powers of 3, like:
30, 31, .... etc

Given that:
Σn=0 to 7 (log3S(n)) = 308, and:

56 ≤ (log3 n=0 to 7 S(n)))≤ 57

Find log3S(14)

 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 2
It hardly matters what the terms of S(n) are.  Since they are powers of 3 and we keep taking logarithms, it's just the exponents that matter.  Let 3^(E(n)) = S(n). If the terms of S(n) are geometric, than the exponents E(n) are arithmetic.

The first given is that the first 8 exponents have sum 308
E(n)=an+b
The sum is 28a+8b
Possibilities: (a,b,E(7))
(1,35,42)
(3,28,49)
(5,21,56)
(7,14,63)
(9,7,70)
If we sum terms 0 through 7 of S(n) the answer isn't much larger than the last term, so its base 3 logarithm isn't much bigger than the last exponent.  So the (5,21,56) is our sequence.

The 14th term of E(n) is therefore 91 = log3S(14)

 Posted by Jer on 2016-02-22 11:03:46 Please log in:

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