The nth term of a geometric sequence is:
T(n) = a*r^(n-1)
where
- T(n) = nth term
- a = first term
- r = common ratio
- n = term number
The first term is given to be a = 3. The seventh term occurs when n = 7, and we're told that T(7) = 1/243.
T(n) = a*r^(n-1)
T(7) = 3*r^(7-1)
1/243 = 3*r^6
1/243 = 3*r^6
r^6 = (1/243)*(1/3)
r^6 = 1/729
r = (1/729)^(1/6)
r = 1/3 or r = -1/3
We have the plus minus because the exponent 6 in r^6 is an even number. It's a similar idea how x^2 = 9 leads to x = 3 or x = -3.
If r = 1/3, then the geometric sequence is:
3, 1, 1/3, 1/9, 1/27, 1/81, 1/243, 1/729, ...
Or if r = -1/3, then the geometric sequence is:
3, -1, 1/3, -1/9, 1/27, -1/81, 1/243, -1/729, ...
I've highlighted the seventh term 1/243 to show it's present in both geometric sequences. Both sequences are nearly identical; however, the second sequence alternates in sign.
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Recall that the infinite geometric sum formula is
S = a/(1-r)
The crucial condition is that -1 < r < 1 must be true for this formula to work.
Luckily r = 1/3 and r = -1/3 fit that interval.
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Let's compute the infinite sum when r = 1/3.
S = a/(1-r)
S = 3/(1-1/3)
S = 3/( 2/3 )
S = 9/2
Therefore, 3+1+1/3+1/9+1/27+1/81+1/243+1/729+... = 9/2
Do the same for the ratio r = -1/3
S = a/(1-r)
S = 3/(1-(-1/3))
S = 3/(1+1/3)
S = 3/( 4/3 )
S = 9/4
Therefore, 3+(-1)+1/3+(-1/9)+1/27+(-1/81)+1/243+(-1/729)+... = 9/4