Posted by Nakum Bakul on Thursday, July 28, 2011 at 7:03am.
The sum of first 13th term as G.P. Is 21 and the sum of 21th term in 13. find the sum of first 34th terms.

I am stuck  math  Reiny, Thursday, July 28, 2011 at 8:19am
a(r^13  1)/(r1) = 21 (#1)
a(r^21  1)/(r1) = 13 (#2)
divide #2 by #1, the "a" will cancel, so will the (r1) to get
(r^21  1)/(r^13  1) = 13/21
I then crossmultiplied and simplified to get
21r^21  13r^13 = 8
At this point I am currently at a standstill.
Perhaps a different approach?
I noticed that 34 = 21 + 13

math  Mgraph, Thursday, July 28, 2011 at 12:59pm
I'll try to show that the equation of Reiny 21r^2113r^13=8 has only one real solution r=1.
Let F(r)=21r^2113r^138, F(1)=0.
F'(r)=441r^20169r^12=
=169r^12(441/169r^81)=169r^2(21/13r^4+1)*
(21/13r^41)=169r^2(21/13r^4+1)* (sqrt(21/13)r^2+1)(sqrt(21/13)r^21)
F'(r)=0 if r=r1=(13/21)^0.25, r=r2=0, or r=r3=(13/25)^0.25
Fmax=F(r1)=1.693443+2.7358<0 Q.E.D.
I am surprised!

math  Reiny, Thursday, July 28, 2011 at 3:09pm
I looked at the situation where r = 1
It satisfies the equation 21r^21  13r^13 = 0
but in the original formula for the sum of a GS, that would make the denominator zero, thus undefined.

math  Mgraph, Thursday, July 28, 2011 at 3:43pm
From the terms of problem => r=1 doesn't satisfy (S13=21, S21=13).
From my proof => such G.P. doesn't exist.
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