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Sum of Infinite Terms of a GP

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Question 50

In the following question, fill in the blanks to make the statements true.

If $36 = 6 \times 6 = 6^{2}, t h e n \frac{1}{36}$ expressed as a power with the base 6 is ___ .

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Solution

Given , $36 = 6 \times 6 = 6^{2}$
So, $\frac{1}{36} = \frac{1}{6 \times 6} = \frac{1}{(6)^{2}} = (6)^{- 2}$

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Sum of Infinite Terms of a GP

Standard X Mathematics

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