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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

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. Check Hiroki's Formula: To determine the correct formula, we need to check if each formula correctly calculates the terms of the sequence when nn is substituted with the term number.
  2. Test Hiroki's Formula: Let's test Hiroki's formula: f(n)=625×(15)nf(n) = 625 \times (\frac{1}{5})^n. We will substitute n=1n=1 to see if we get the first term of the sequence, which is 125125.\newlinef(1)=625×(15)1=625×15=125f(1) = 625 \times (\frac{1}{5})^1 = 625 \times \frac{1}{5} = 125.\newlineThis matches the first term of the sequence.
  3. Test Mapiya's Formula: Now let's test Mapiya's formula: f(n)=125×(15)nf(n) = 125 \times (\frac{1}{5})^n. We will also substitute n=1n=1 to see if we get the first term of the sequence.f(1)=125×(15)1=125×15=25.f(1) = 125 \times (\frac{1}{5})^1 = 125 \times \frac{1}{5} = 25.This does not match the first term of the sequence; it matches the second term instead.
  4. Conclusion: Since Hiroki's formula correctly produced the first term of the sequence when n=1n=1, and Mapiya's formula did not, we can conclude that only Hiroki's formula is correct.

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