CAT 2025

Slot 1

QA

Question & SolutionQues & Sol

In the set of consecutive odd numbers $\left\{1,3,5,...,57\right\}$, there is a number k such that the sum of all the elements less than k is equal to the sum of all the elements greater than k . Then, k equals

1. Concept Used

• Topic: Number Series — Sum of Consecutive Odd Numbers
• Formula: $\text{Sum of first } n \text{ odd numbers} = n^2$

2. Calculation

The set is ${1, 3, 5, ..., 57}$. The number of terms = $\frac{57 - 1}{2} + 1 = 29$.

The total sum of all 29 odd numbers is: $$S_{\text{total}} = 29^2 = 841$$

Let $k$ be the $m^{th}$ term of the sequence. Since these are consecutive odd numbers, the $m^{th}$ term is $2m - 1$.

The sum of the first $(m-1)$ odd numbers (all elements less than $k$) is: $$S_{\text{less}} = (m-1)^2$$

The sum of all elements greater than $k$ is: $$S_{\text{greater}} = S_{\text{total}} - k - S_{\text{less}} = 841 - (2m-1) - (m-1)^2$$

Setting $S_{\text{less}} = S_{\text{greater}}$: $$(m-1)^2 = 841 - (2m-1) - (m-1)^2$$

$$2(m-1)^2 = 841 - (2m - 1)$$

$$2(m^2 - 2m + 1) = 842 - 2m$$

$$2m^2 - 4m + 2 = 842 - 2m$$

$$2m^2 - 2m - 840 = 0$$

$$m^2 - m - 420 = 0$$

Factoring the quadratic: $$(m - 21)(m + 20) = 0$$

Since $m$ must be positive: $m = 21$.

The 21st odd number is: $$k = 2(21) - 1 = 41$$

Verification: Sum of first 20 odd numbers $= 20^2 = 400$. Sum of remaining 8 odd numbers (43 to 57) $= 29^2 - 20^2 - 41 = 841 - 400 - 41 = 400$. ✅

3. Solution

Answer = Option 1 (41) ✅

The final calculated value is k = 41, which is the 21st term of the sequence where the sum of all elements less than it equals the sum of all elements greater than it (both equal 400).