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{:[{[f(1)=25],[f(n)=f(n-1)*(-(1)/(5))]:}],[f(3)=◻]:}

{f(1)=25f(n)=f(n−1)⋅(−15) \left\{\begin{array}{l}f(1)=25 \\ f(n)=f(n-1) \cdot\left(-\frac{1}{5}\right)\end{array}\right. \newlinef(3)= f(3)=

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Q. {f(1)=25f(n)=f(n−1)⋅(−15) \left\{\begin{array}{l}f(1)=25 \\ f(n)=f(n-1) \cdot\left(-\frac{1}{5}\right)\end{array}\right. \newlinef(3)= f(3)=
  1. Given initial condition and recursive formula: We are given the initial condition and the recursive formula for the function ff:f(1)=25f(1) = 25f(n)=f(n−1)⋅(−15)f(n) = f(n-1) \cdot \left(-\frac{1}{5}\right)To find f(3)f(3), we first need to find f(2)f(2) using the recursive formula.
  2. Finding f(2)f(2) using the recursive formula: Using the recursive formula, we calculate f(2)f(2):
    f(2)=f(1)⋅(−15)f(2) = f(1) \cdot \left(-\frac{1}{5}\right)
    f(2)=25⋅(−15)f(2) = 25 \cdot \left(-\frac{1}{5}\right)
    f(2)=−5f(2) = -5
    We have found the value of f(2)f(2) to be −5-5.
  3. Calculating f(2)f(2): Now we can use the value of f(2)f(2) to find f(3)f(3):\newlinef(3)=f(2)⋅(−15)f(3) = f(2) \cdot \left(-\frac{1}{5}\right)\newlinef(3)=−5⋅(−15)f(3) = -5 \cdot \left(-\frac{1}{5}\right)\newlinef(3)=1f(3) = 1\newlineWe have found the value of f(3)f(3) to be 11.

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