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Shrinking payments

CalculusEasy

A contract pays $6$ dollars at the end of year $1$ , and every later payment is $\frac{3}{4}$ of the one before, forever. Ignoring interest, what is the total of all the payments, in dollars? Give the answer as an integer.

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Hint

The payments form a geometric series; you need its first term and its ratio.

Solution

The payments are $6$ , $6 \times \frac{3}{4}$ , $6 \times (\frac{3}{4})^{2}$ and so on.
A geometric series with first term $a$ and ratio $∣ r ∣< 1$ sums to $\frac{a}{1 - r}$ .
Here that is $\frac{6}{1 - \frac{3}{4}} = \frac{6}{\frac{1}{4}} = 24$ .

Answer: $24$

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