In the set of consecutive odd numbers {1, 3, 5, ….., 57}, there is a number of 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.
Correct Answer :
41
Solution :
The correct answer is 41.
First, let's look at the given set of consecutive odd numbers: {1, 3, 5, ..., 57}. This forms an arithmetic progression where the first term is a = 1 and the common difference is d = 2.
We can find the total number of terms, let's call it n, in this sequence. The formula for the n-th term of an arithmetic progression is given by:
Setting the last term to 57, we have:
So, there are 29 odd numbers in total.
We know that the sum of the first n consecutive odd numbers starting from 1 is exactly n squared. The total sum of all 29 numbers in the set is therefore:
Let the unknown number k be the m-th term in the sequence. This means:
We are given that the sum of the numbers less than k equals the sum of the numbers greater than k.
The numbers strictly less than k are the first m - 1 terms of the sequence. Using our rule for the sum of odd numbers, their sum is:
The sum of the numbers strictly greater than k is the total sum of all terms minus the sum of the first m terms (which includes k itself). This can be expressed as:
Equating the two sums as specified in the problem, we get:
Expanding the left side:
Rearranging the equation to form a standard quadratic equation:
Dividing the entire equation by 2 to simplify:
Now, we factor the quadratic equation. We look for two numbers that multiply to -420 and add to -1. These numbers are -21 and 20.
Since m represents the position of the term in the sequence, it must be a positive integer. Therefore, we can discard -20, meaning m = 21.
Finally, we can find the value of k since we now know it is the 21st term in the sequence:
Thus, the value of k is 41.
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