CAT 2025Slot 1QAQuestion & Solution
Question
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
Options
41
39
43
37
Solution
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 (21^{st}) 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).