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Hiroki and Mapiya were asked to find an explicit formula for the sequence 125,25,5,1,dots, where the first term should be f(1). Hiroki said the formula is f(n)=625*((1)/(5))^(n), and Mapiya said the formula is f(n)=125*((1)/(5))^(n). Which one of them is right? Choose 1 answer: (A) Only Hiroki (B) Only Mapiya (c) Both Hiroki and Mapiya (D) Neither Hiroki nor Mapiya

Hiroki and Mapiya were asked to find an explicit formula for the sequence \newline125,25,5,1,…125, 25, 5, 1, \dots, where the first term should be \newlinef(1)f(1).\newlineHiroki said the formula is \newlinef(n)=625⋅(15)nf(n) = 625 \cdot \left(\frac{1}{5}\right)^n, and\newlineMapiya said the formula is \newlinef(n)=125⋅(15)nf(n) = 125 \cdot \left(\frac{1}{5}\right)^n.\newlineWhich one of them is right?\newlineChoose 11 answer:\newline(A) Only Hiroki\newline(B) Only Mapiya\newline(C) Both Hiroki and Mapiya\newline(D) Neither Hiroki nor Mapiya

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Full solution

Q. Hiroki and Mapiya were asked to find an explicit formula for the sequence \newline125,25,5,1,…125, 25, 5, 1, \dots, where the first term should be \newlinef(1)f(1).\newlineHiroki said the formula is \newlinef(n)=625⋅(15)nf(n) = 625 \cdot \left(\frac{1}{5}\right)^n, and\newlineMapiya said the formula is \newlinef(n)=125⋅(15)nf(n) = 125 \cdot \left(\frac{1}{5}\right)^n.\newlineWhich one of them is right?\newlineChoose 11 answer:\newline(A) Only Hiroki\newline(B) Only Mapiya\newline(C) Both Hiroki and Mapiya\newline(D) Neither Hiroki nor Mapiya
  1. Identify the pattern: We have the sequence: 125,25,5,1,…125, 25, 5, 1, \ldots\newlineIdentify the pattern in the sequence.
  2. Determine first term and common ratio: The sequence is geometric because each term is obtained by multiplying the previous term by a common ratio.\newlineDetermine the first term (f(1)f(1)) and the common ratio (rr).\newlineFirst term: f(1)=125f(1) = 125\newlineCommon ratio: r=25125=15r = \frac{25}{125} = \frac{1}{5}
  3. Test Hiroki's formula: Now, let's check Hiroki's formula: f(n)=625×(15)nf(n) = 625 \times \left(\frac{1}{5}\right)^n\newlineTest Hiroki's formula with n=1n = 1 to see if it gives the first term of the sequence.\newlinef(1)=625×(15)1=625×15=125f(1) = 625 \times \left(\frac{1}{5}\right)^1 = 625 \times \frac{1}{5} = 125\newlineHiroki's formula gives the correct first term.
  4. Test Mapiya's formula: Next, let's check Mapiya's formula: f(n)=125⋅(15)nf(n) = 125 \cdot \left(\frac{1}{5}\right)^n\newlineTest Mapiya's formula with n=1n = 1 to see if it gives the first term of the sequence.\newlinef(1)=125⋅(15)1=125⋅15=25f(1) = 125 \cdot \left(\frac{1}{5}\right)^1 = 125 \cdot \frac{1}{5} = 25\newlineMapiya's formula does not give the correct first term.
  5. Conclusion: Since Hiroki's formula gives the correct first term and follows the pattern of the sequence, Hiroki is correct.\newlineMapiya's formula does not give the correct first term, so Mapiya is incorrect.

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