MCQ
The product of $(64)(64)^{1 / 7}(64)^{1 / 49} \ldots \infty$ is equal to-
- A64
- B32
- ✓128
- D256
✓
Answer
Correct option: C.
128
(c) 128
Explanation:
Here, (64)(64) $\frac{1}{7}(64) \frac{1}{49} \ldots \infty$
$\left(1+\frac{1}{7}+\frac{1}{49}+\ldots \infty\right)$
Since $1+\frac{1}{7}+\frac{1}{49}+\ldots \infty$ is in G.P.
Here, $a=1, r=\frac{1}{7} \div 1=\frac{1}{7}$
$S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\frac{1}{7}}=\frac{7}{6}$
So, from equation (i),
$\begin{array}{l}64^{\left(1+\frac{1}{7}+\frac{1}{49}+\ldots \infty\right)}={ }_{64^{\frac{7}{6}}} \\ ={ }_{(2)}^{66 \frac{7}{6}} \\ =2^7 \\ =128\end{array}$
Explanation:
Here, (64)(64) $\frac{1}{7}(64) \frac{1}{49} \ldots \infty$
$\left(1+\frac{1}{7}+\frac{1}{49}+\ldots \infty\right)$
Since $1+\frac{1}{7}+\frac{1}{49}+\ldots \infty$ is in G.P.
Here, $a=1, r=\frac{1}{7} \div 1=\frac{1}{7}$
$S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\frac{1}{7}}=\frac{7}{6}$
So, from equation (i),
$\begin{array}{l}64^{\left(1+\frac{1}{7}+\frac{1}{49}+\ldots \infty\right)}={ }_{64^{\frac{7}{6}}} \\ ={ }_{(2)}^{66 \frac{7}{6}} \\ =2^7 \\ =128\end{array}$
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