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Exercise 5.4 · Q3

Q.Using a geometric series, write 0.175175175175…0.175175175175\ldots as a fraction.

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✓ Free question

Writing the repeating decimal as an infinite GP with a=1751000a=\dfrac{175}{1000} and r=11000r=\dfrac{1}{1000} gives the fraction 175999\dfrac{175}{999}.

[!FORMULA] S∞=a1−rS_\infty=\dfrac{a}{1-r}, for ∣r∣<1|r|<1

aa = first term, rr = common ratio.

  1. Split the repeating decimal into an infinite sum: 0.175175175…=1751000+17510002+17510003+⋯0.175175175\ldots=\dfrac{175}{1000}+\dfrac{175}{1000^2}+\dfrac{175}{1000^3}+\cdots, a GP with a=1751000a=\dfrac{175}{1000} and r=11000r=\dfrac{1}{1000} (each block of 3 digits is 11000\dfrac{1}{1000} of the previous).
  2. Apply the infinite-GP sum formula: S∞=a1−r=17510001−11000=17510009991000S_\infty=\dfrac{a}{1-r}=\dfrac{\frac{175}{1000}}{1-\frac{1}{1000}}=\dfrac{\frac{175}{1000}}{\frac{999}{1000}}.
  3. The 11000\dfrac{1}{1000} cancels: S∞=175999S_\infty=\dfrac{175}{999}.
  4. Check lowest terms: 175=52×7175=5^2\times7 and 999=33×37999=3^3\times37 share no common factor, so 175999\dfrac{175}{999} is already fully reduced.
  5. Sanity check: 175÷999≈0.175175…175\div999\approx0.175175\ldots ✓, matching the given decimal.
✓Final answer

175999\dfrac{175}{999}.

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