CAT 2025 Slot 1 QA Question 9

First, We analyze the set we're dealing with. It's the set of consecutive odd numbers starting from 1 up to 57. So, that would be 1, 3, 5, ..., 57. consecutive odd numbers form an arithmetic sequence where each term increases by 2.

Let me recall the formula for the nth term of an arithmetic sequence. It's a_n = a_1 + (n - 1)d, where a_1 is the first term, d is the common difference, and n is the number of terms. Here, a_1 is 1, d is 2, and a_n is 57. So, plugging in the values:

57 = 1 + (n - 1)2 57 - 1 = (n - 1)2 56 = 2(n - 1) n - 1 = 28 n = 29

So, there are 29 terms in this sequence. That means the number k we're looking for is somewhere in the middle of this sequence. Since 29 is odd, the middle term is the 15th term. Calculate that.

The 15th term would be a_15 = 1 + (15 - 1)*2 = 1 + 28 = 29. Wait, but 29 isn't one of the options. Hmm, maybe I made a mistake. Let me check again.

Wait, no, the options are 37, 39, 41, 43. So, 29 isn't an option. Maybe I misunderstood the problem. It says the sum of elements less than k equals the sum of elements greater than k. So, maybe k isn't necessarily the middle term, but a term where the sum on both sides is equal.

Let me think about the total sum of the sequence. If I can find the total sum, then the sum less than k plus the sum greater than k would be equal to the total sum minus k. Since the sum less than k equals the sum greater than k, each of those sums would be (Total Sum - k)/2.

So, first, let me find the total sum of the sequence. The formula for the sum of an arithmetic sequence is S = n/2 * (a_1 + a_n). Plugging in the values:

S = 29/2 (1 + 57) = 29/2 58 = 29 * 29 = 841.

Wait, 29 times 29 is 841? Let me verify: 3030 is 900, so 2929 is 841. Yes, that's correct.

So, the total sum is 841. Now, the sum less than k is equal to the sum greater than k. Let the sum less than k as S_less and the sum greater than k as S_greater. So, S_less = S_greater.

But S_less + S_greater + k = 841. Since S_less = S_greater, then 2*S_less + k = 841. Therefore, S_less = (841 - k)/2.

So, We find k such that the sum of all terms less than k is equal to (841 - k)/2.

Now, We find the position of k in the sequence. Let the number of terms less than k as m. Since the sequence is 1, 3, 5, ..., 57, each term can be represented as 2i - 1, where i is the term number. So, k = 2i - 1 for some i.

The sum of the first m terms is S_less = m/2 [21 + (m - 1)2] = m/2 [2 + 2m - 2] = m/2 * 2m = m^2.

Wait, that's interesting. The sum of the first m odd numbers is m^2. So, S_less = m^2.

But S_less is also equal to (841 - k)/2. So, m^2 = (841 - k)/2.

But k is the (m + 1)th term, right? Because if we have m terms less than k, then k is the (m + 1)th term. So, k = 2(m + 1) - 1 = 2m + 1.

So, substituting k into the equation:

m^2 = (841 - (2m + 1))/2 m^2 = (840 - 2m)/2 m^2 = 420 - m m^2 + m - 420 = 0

Now, I need to solve this quadratic equation: m^2 + m - 420 = 0.

Using the quadratic formula: m = [-b ± sqrt(b^2 - 4ac)]/(2a). Here, a = 1, b = 1, c = -420.

So, discriminant D = 1 + 1680 = 1681. \ is 41.

So, m = [-1 ± 41]/2. Since m can't be negative, m = (40)/2 = 20.

So, m = 20. Therefore, k is the 21st term. Let me find the 21st term.

k = 2*21 - 1 = 42 - 1 = 41.

So, k is 41, which is option A.

If k is 41, then the sum of terms less than 41 is the sum of the first 20 terms. The sum of the first 20 odd numbers is 20^2 = 400. The sum of the terms greater than 41 would be the total sum minus 41 minus 400. So, 841 - 41 - 400 = 841 - 441 = 400. Yes, that matches. So, the sum less than k is 400, and the sum greater than k is also 400. So, k is indeed 41.

I think that makes sense