In the set of consecutive odd numbers $\{1, 3, 5, \ldots, 57\}$ , 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

✅ Correct Option: 1

The set $\{1, 3, 5, \ldots, 57\}$ contains all odd numbers from 1 to 57.

The $n$ -th odd number is $2n - 1$ , so:

$2n - 1 = 57$

$n = 29$

There are 29 terms in the set.

The sum of first $n$ odd numbers $= n^2$

For example, $1 + 3 + 5 = 9 = 3^2$ and $1 + 3 + 5 + 7 = 16 = 4^2$

Total sum of all 29 odd numbers:

$S = 29^2 = 841$

Let $k$ be the $m$ -th odd number in the sequence, so $k = 2m - 1$ .

Sum of all elements less than $k$ = Sum of first $(m-1)$ odd numbers $= (m-1)^2$

Sum of all elements greater than $k$ $= 841 - (m-1)^2 - (2m-1)$

Setting both sums equal:

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

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

$(m-1)^2 = 421 - m$

$m^2 - 2m + 1 = 421 - m$

$m^2 - m - 420 = 0$

$(m - 21)(m + 20) = 0$

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

$k = 2m - 1$

$k = 2(21) - 1$

$k = 41$

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

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CAT Past Year Questions (Topic-Wise)

In the set of consecutive odd numbers $\{1, 3, 5, \ldots, 57\}$ , 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