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Question 59

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

Solution

The sum of all the elements in the given set = Sum of first 29 odd numbers = $$29^2$$ = 841

Let's assume that k is the $$m_{th}$$ term. Sum of terms less than k = sum of first (m-1) odd numbers = $$(m-1)^2$$

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

$$841 - m^2 = m^2 - 2m + 1$$

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

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

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

m = 21 -20

m = 20. And the 20th term is 2*m+1 = 41

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