To determine the first five terms of the sequence defined by the formula \( a_n = 112 \left( \frac{1}{4} \right)^{n-1} \), we will evaluate this formula for \( n = 1, 2, 3, 4, \) and \( 5 \).
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For \( n = 1 \): \[ a_1 = 112 \left( \frac{1}{4} \right)^{1-1} = 112 \left( \frac{1}{4} \right)^0 = 112 \cdot 1 = 112 \]
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For \( n = 2 \): \[ a_2 = 112 \left( \frac{1}{4} \right)^{2-1} = 112 \left( \frac{1}{4} \right)^1 = 112 \cdot \frac{1}{4} = 28 \]
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For \( n = 3 \): \[ a_3 = 112 \left( \frac{1}{4} \right)^{3-1} = 112 \left( \frac{1}{4} \right)^2 = 112 \cdot \frac{1}{16} = 7 \]
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For \( n = 4 \): \[ a_4 = 112 \left( \frac{1}{4} \right)^{4-1} = 112 \left( \frac{1}{4} \right)^3 = 112 \cdot \frac{1}{64} = \frac{7}{4} \]
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For \( n = 5 \): \[ a_5 = 112 \left( \frac{1}{4} \right)^{5-1} = 112 \left( \frac{1}{4} \right)^4 = 112 \cdot \frac{1}{256} = \frac{7}{16} \]
Now, compiling the first five terms, we have:
\[ 112, 28, 7, \frac{7}{4}, \frac{7}{16} \]
Thus, the correct response is:
112, 28, 7, \(\frac{7}{4}\), \(\frac{7}{16}\).