Question

For a geometric sequence, a1 = -2 r=2 and an= -64. Find n and Sn The n's are subscript, but I don't know how to type them like that on here.

Answer 1

A geometric sequence is much like a geometric series in that the formula is as follows:
\[a_{n}=ar^{n}\] when n starts at 0 or\[a_{n}=ar^{n-1}\] when n starts at 1. Since you are given \[a_{1} = -2\] Then we will use the \[a_{n}=ar^{n-1}\]formula because that starts at n=1, so n can be deduced by plugging the values given in the formula, i.e.\[a_{1}=-2=ar^{1-1}=ar^{0}=a\]\[a_{n}=-64=-2*r^{n-1}\]\[-64/-2=2^{n-1}=2^{n}2^{-1}\]\[64/2=2^{n}/2\]\[64=2^{n}\]\[\log_{2}(64)=n=6\]So this saya that the last value in the sequence is -64 and that term is n=6.

Sn is in the next post

Answer 2

From above, \[s_{n}=ar^{1-1}+ar^{2-1}+ar^{3-1}+...+ar^{n-1}\]
\[s_{n}=a+ar+ar^{2}+...+ar^{n-1}\]
Now if we multiply r on both sides of this equation, we get
\[rs_{n}=ar+ar^{2}+ar^{3}+...+ar^{n-1}r\]
The last term is \[ar^{n-1}r=ar^{n-1+1}=ar^{n}\]
Subtract rSn from Sn
\[s_{n}-rs_{n}=a+ar+ar^{2}+...+ar^{n-1} -\]
\[ (ar+ar^{2}+ar^{3}+...+ar^{n})\]
Notice that all the terms cancel each other except the first and the last.
\[s_{n}-rs_{n}=a-ar^{n}\]\[s_{n}(1-r)=a(1-r^{n})\]\[s_{n}=a(1-r^{n})/(1-r)=-2(1-2^{n})/(-1)=2(1-2^{n})\]Knowing that a1=a