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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 = \(\rm 144^{\tfrac{1}{3}+\tfrac{1}{9}+\tfrac{1}{27}+\ ... \infty}=144^{\tfrac{1}{2}}=\sqrt144 = 12\).

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