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)=441r^20-169r^12=

=169r^12(441/169r^8-1)=169r^2(21/13r^4+1)*

(21/13r^4-1)=169r^2(21/13r^4+1)* (sqrt(21/13)r^2+1)(sqrt(21/13)r^2-1)

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+2.735-8<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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