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

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