# The value of $\rm 144^{\tfrac{1}{3}} 144^{\tfrac{1}{9}} 144^{\tfrac{1}{27}}\ ...\ \infty$ is:

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The value of $\rm 144^{\tfrac{1}{3}} 144^{\tfrac{1}{9}} 144^{\tfrac{1}{27}}\ ...\ \infty$ is:
1. 4
2. 12
3. 11
4. 9

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Correct Answer - Option 2 : 12

Concept:

Geometric Progression (GP):

• The series of numbers where the ratio of any two consecutive terms is the same is called a Geometric Progression.
• A Geometric Progression of n terms with first term a and common ratio r is represented as:

a, ar, ar2, ar3, ..., arn-2, arn-1.

• The sum of the first n terms of a GP is: Sn = $\rm a\left(\frac{r^n-1}{r-1}\right)$.
• The sum to ∞ of a GP, when |r| < 1, is: S = $\rm \frac{a}{1-r}$.

Calculation:

$\rm 144^{(\tfrac{1}{3}+\tfrac{1}{9}+\tfrac{1}{27}............\infty)}$

Let us consider the infinite series $\rm \dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ ... \infty$.

Here, a = $\rm \dfrac{1}{3}$ and r = $\rm \dfrac{\tfrac{1}{9}}{\tfrac{1}{3}}=\dfrac{1}{3}$.

∴ S = $\rm \dfrac{a}{1-r}=\dfrac{\tfrac{1}{3}}{1-\tfrac{1}{3}}=\dfrac{\tfrac{1}{3}}{\tfrac{2}{3}}=\dfrac{1}{2}$.

Now, let P = $\rm 144^{(\tfrac{1}{3}+\tfrac{1}{9}+\tfrac{1}{27}............\infty)}$

∴ P = $\rm 144^{\tfrac{1}{3}+\tfrac{1}{9}+\tfrac{1}{27}+\ ... \infty}=144^{\tfrac{1}{2}}=\sqrt144 = 12$.