For any natural number k , let a_k=3^(k) . The smallest natural number m for which \ (a_1 )¹× (a_2 )²× cdots × (a_20 )²⁰ \< a_21timesa_22× cdots timesa_20+m \ is

CAT 2025 Slot 1 · Quant · Question 19

CAT 2025 Slot 1 previous year question on Surds and Indices, with the answer and a step-by-step solution.

Surds and IndicesAlgebraMultiple choiceMedium

For any natural number k, let {a}_{k}={3}^{k}. The smallest natural number m for which \left\{{\left({a}_{1}\right)}^{1}\times{\left({a}_{2}\right)}^{2}\times \cdots \times{\left({a}_{20}\right)}^{20}\right\}<\left\{{a}_{21}\times{a}_{22}\times \cdots \times{a}_{20+m}\right\} is

A
59
B
56
C
58
D
57

Pick an option, then check.

Answer & solution

Answer: Option C

Everything is a power of 3, so compare exponents only.

Left side: {\left({3}^{k}\right)}^{k}={3}^{{k}^{2}},

so the exponent is \sum_{k=1}^{20}{{k}^{2}}=\frac{n(n+1)(2n+1)}{6}=\frac{20\cdot 21\cdot 41}{6}=2870.

Right side: the exponents are 21,22,\ldots ,20+m,

an AP of m terms, summing to \frac{m(a+l)}{2}=\frac{m\left( 21+20+m \right)}{2}=\frac{m\left( 41+m \right)}{2}.

Since the base 3 exceeds 1, the inequality holds exactly when 2870<\frac{m\left( 41+m \right)}{2},

that is {m}^{2}+41m>5740.

At m=57:

3249+2337=5586, which is too small.

At m=58:

3364+2378=5742, which clears it.

So the smallest value is m=58.

CAT 2025 Slot 1 · Quant · Question 19

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