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Hi good people!..juriemagic here, lovely day is it not????..oki-doki, lets get to this VERY small problem..laugh

 

It is given: T1=a, T2=ar, T3=ar^2 and T4=ar^3

Also, S1=T1, S2=T1+T2..and so on

 

There a numerous questions on this which I have solved, However..two questions remain, If someone would be so kind to please embarrass me?.

 

Give an expression for r x S6, and

show that (1-r)S6=a(1-r^6).

 

all and any help as usual will be very much appreciated!. thank you all..

 Nov 12, 2015

Best Answer 

 #1
avatar+33666 
+10

This is a geometric progression.  Here's a derivation for the n'th term:

 

 geometric progression

 Nov 12, 2015
 #1
avatar+33666 
+10
Best Answer

This is a geometric progression.  Here's a derivation for the n'th term:

 

 geometric progression

Alan Nov 12, 2015
 #2
avatar+26403 
+10

Hi good people!..juriemagic here, lovely day is it not????..oki-doki, lets get to this VERY small problem..

 

It is given: T1=a, T2=ar, T3=ar^2 and T4=ar^3

Also, S1=T1, S2=T1+T2..and so on

 

Give an expression for r x S6, and

show that (1-r)S6=a(1-r^6).

 

\(\begin{array}{rcll} s_n &=& a + ar + ar^2+ \dots + ar^{n-2} + ar^{n-1} \\ r\cdot s_n &=& ar + ar^2+ \dots + ar^{n-1} + ar^{n} \\ \hline s_n - r\cdot (s_n) &=& a - ar^{n} \\ s_n\cdot (1-r) &=& a\cdot ( 1-r^n ) \\ s_n &=& a\cdot \frac{1-r^n}{1-r} \\ \hline s_6 &=& a\cdot \frac{1-r^6}{1-r} \\ r\cdot s_6 &=& r\cdot a\cdot \frac{1-r^6}{1-r}\\\\ (1-r)\cdot s_6 &=& (1-r)\cdot a\cdot \frac{1-r^6}{1-r}\\ &=& a\cdot (1-r^6) \end{array}\)

 

laugh

 Nov 12, 2015
 #3
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0

Two different aproaches, thank you Alan and Heureka!..I do appreciate!!..Have a blessed day!

 Nov 12, 2015

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