CAT 2025

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Exponent Laws, Summation of Series (Sum of Squares and Sum of Natural Numbers)
• Formula: $$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$

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2. Calculation

Since $a_k = 3^k$, we substitute to simplify both sides of the inequality.

Left Hand Side (LHS):

$$(a_1)^1 \times (a_2)^2 \times \cdots \times (a_{20})^{20} = (3^1)^1 \times (3^2)^2 \times \cdots \times (3^{20})^{20}$$

$$= 3^{1^2} \times 3^{2^2} \times \cdots \times 3^{20^2} = 3^{\sum_{k=1}^{20} k^2}$$

Using the sum of squares formula: $$\sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870$$

So, $\text{LHS} = 3^{2870}$.

Right Hand Side (RHS):

$$a_{21} \times a_{22} \times \cdots \times a_{20+m} = 3^{21} \times 3^{22} \times \cdots \times 3^{20+m} = 3^{\sum_{k=21}^{20+m} k}$$

The exponent on the RHS is: $$\sum_{k=21}^{20+m} k = \sum_{k=1}^{20+m} k - \sum_{k=1}^{20} k = \frac{(20+m)(21+m)}{2} - \frac{20 \cdot 21}{2}$$

$$= \frac{(20+m)(21+m) - 420}{2} = \frac{m^2 + 41m + 420 - 420}{2} = \frac{m^2 + 41m}{2}$$

Setting up the inequality:

Since the bases are equal (both are powers of 3), we compare the exponents:

$$2870 < \frac{m^2 + 41m}{2} \implies 5740 < m^2 + 41m \implies 5740 < m(m + 41)$$

Now we test the options to find the smallest $m$:

• $m = 56$: $56 \times 97 = 5432 < 5740$ ❌
• $m = 57$: $57 \times 98 = 5586 < 5740$ ❌
• $m = 58$: $58 \times 99 = 5742 > 5740$ ✅

So $m = 58$ is the smallest natural number satisfying the inequality.

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3. Solution

Answer = Option 1 (58) ✅

The smallest natural number $m$ for which the inequality holds is $m = 58$, since $58 \times 99 = 5742 > 5740$.