math
posted by Nakum Bakul .
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.

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 
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! 
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. 
From the terms of problem => r=1 doesn't satisfy (S13=21, S21=13).
From my proof => such G.P. doesn't exist.
Respond to this Question
Similar Questions

math
1) If the sum of the first n terms of a series is Sn where Sn=2n²n, (a)prove that the series is an AP, stating the first term and the common difference; (b)find the sum of the terms from the 3rd to the 12th inclusive 2) In an AP … 
Math
1) If the sum of the first n terms of a series is Sn where Sn=2n²n, (a)prove that the series is an AP, stating the first term and the common difference; (b)find the sum of the terms from the 3rd to the 12th inclusive 2) In an AP … 
Geometric
1) The sum of last three terms of GP having n terms is 1024 time the sume of first 3 terms of GP. If 3rd term is 5. Find the last term. 2) The sum of the first eight terms of a GP is five times the sum of the first four terms. Find … 
maths
1) The first term of arithmetic progression is 20 and the sum of it's term is 250. find it's last term if the number of it's terms is 10 2) Find the fifth term from the arithmetic progression 12, 9, 6 hence find the sum of it's … 
Math
Sum of4th and the 6th terms of an a,p is 42 . While the sum of the 3rd and 9th terms of the same a.p is 52. Find the (a) first term (b)common difference (c) sum of the first ten terms of the progression. Solution 
Further mathematics
An exponential sequence (GP)of positive terms and a linear sequence (AP)have the same first term.the sum of their first terms is 3,the sum of their second terms is 3 and the sum of their third terms is 6. Find the sum of their fifth … 
math
The third term of a G.P is 24 and the sum of the first two terms is 288. If all the terms are positive,find a)the first term and the common ratio b)the sum of all the terms from 4th term to the 8th term 
arithmetic
in an arithmetic progression the 13th term is 27 and the 7th term is three times the second term find;the common difference,the first term and the sum of the first ten terms 
mathematics
the 10th term of an arithmetic sequence is equal to the sum of 40 and 5th term.If the 15th term is 127 what is its common difference? 
Math3c
1.Find an A.P. such that the sum of the first three terms is one half the sum of the next four terms,the first term being 12. 2.In an A.P.,the sum of the first three terms is 18,and the sum of the squares of these terms is 126.Find …