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Find the value of A1 for an infinite geometric series with S = -45 and R=-1/9
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Please show YOUR work. Do you have a formula for such a sum? Maybe: \(S = \dfrac{A_{1}}{1-r}\)? Please substitute known values and solve for the remaining value.
-45=A1/1+.111111 A1=-45*1+.11111 A1=-44????
tkhunny
-50
Try again. Please be more careful with your notation A1/1 - b = (a1/1) - b You mean A1/(1 - b) and don't use a calculator. Keep the fractions.
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the computer doesnt accept the fractions it tells me to change to decimal
Use a piece of paper. Format the answer as required.
1 + 1/9 = 9/9 + 1/9 = 10/9 You don't need a calculator for that.
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