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Find a general formula for the nth term of the geometric series with: a6=1 and r=1/5
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\(\Large a_{\color{brown}{ n}}=a_1r^{{\color{brown}{ n}}-1}\) does that ring a bell?
A(n) = (6^5)(1/5)^(n-1)
yes no no the foggiest just posting for kicks my cat typed it in wrongly copied and pasted ?
@jdoe0001 yea thanks
@alyssa_nicole thats not an option :-( there is A(n) = (5^6)(1/5)^(n-1) though, is that what you meant?
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yess.
ok well... we know what the 6th term is , \(\bf a_6 = 1\) and we know what "r" is, or the common ratio, \(\bf r = \frac{1}{5}\) so... what's the 1st term? well.. we dunno but we know the 6th term, and the common ratio thus
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