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)/(r-1) = 21 (#1)
a(r^21 - 1)/(r-1) = 13 (#2)
divide #2 by #1, the "a" will cancel, so will the (r-1) to get
(r^21 - 1)/(r^13 - 1) = 13/21
I then cross-multiplied and simplified to get
21r^21 - 13r^13 = 8
At this point I am currently at a stand-still.
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^21-13r^13=8 has only one real solution r=1.
Let F(r)=21r^21-13r^13-8, F(1)=0.
F'(r)=0 if r=r1=-(13/21)^0.25, r=r2=0, or r=r3=(13/25)^0.25
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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