Question 56

Consider the set of numbers {1, 3, $$3^{2}$$, $$3^{3}$$,…...,$$3^{100}$$}. The ratio of the last number and the sum of the remaining numbers is closest to:

Solution

Set : {$$1, 3, 3^{2}, 3^{2},…...,3^{100}$$}

Clearly, this set is a G.P. with common ratio, $$r = 3$$

Sum of G.P. = $$\frac{a (r^n - 1)}{r - 1}$$

Number of terms = 101

Last term = $$3^{100}$$

Sum of remaining terms = $$\frac{1 (3^{100} - 1)}{3 - 1}$$

= $$\frac{3^{100} - 1}{2}$$

$$\therefore$$ Required ratio = $$\frac{3^{100}}{\frac{3^{100} - 1}{2}}$$

= $$\frac{3^{100} \times 2}{3^{100} - 1}$$

$$\approx \frac{3^{100} \times 2}{3^{100}} = 2$$

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