Question 969860: Find the sum.
(3/8)+(3/4)+(3/2)+3+...+192 Found 3 solutions by stanbon, rothauserc, MathTherapy:Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website! Find the sum.
(3/8)+(3/4)+(3/2)+3+...+192
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a = (3/8)
r = (3/4)/(3/8) = 2
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Solve::
a*r^n = 192
r^n = 192/(3/8) = 512
2^n = 512
n = 9
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Ans: S= a[r^n -1]/(r-1)
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S(9) = (3/8)[2^9 - 1]/[2-1] = (3/8)[511] = 191 5/8
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Cheers,
Stan H.
----------------- Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website! (3/8)+(3/4)+(3/2)+3+...+192
this is a geometric series, the formula for the nth term in a geometric series is
xn = ar^(n-1) where a is the first term and r is the common ratio
in this case a = (3/8) and r = 2, we want to find n for the term 192
192 = (3/8) * 2^(n-1)
2^(n-1) = 512
n = 10
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now we want to find the sum of the first 10 terms of our geometric series
sum = a * ((1-r^n) / (1-r)) = (3/8)*((1-(2^10)) / (1-2)) = 383.625