Question

The sum of 444 consecutive odd numbers is 404040.
What is the second number in this sequence?

Answer 1

The solution of your question is given in the pic above!!!

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Answer 2

Let the series of the odd numbers be $\sf{a_n=1+2(n-1)}$.

If the sequence starts from 'α'th odd number and ends in 'β'th odd number, the sum of the series would be $\sf{S_{\beta } -S_{\alpha }}$.

But there are 444 numbers. The number of the numbers from α to β can be equated.

$\implies\sf{\beta -\alpha +1=444}$

$\therefore\sf{\beta =\alpha +443}$

We found that the sum of the series will be $\sf{S_{\alpha +433}-S_{\alpha }}$.

The sum of the first n odd numbers is the following.

$\sf{S_n=\dfrac{n(a+l)}{2} =\dfrac{n\{2+2(n-1)\}}{2} =\dfrac{2n^2}{2} =n^2}$

Therefore, we can equate the series sum and 404040.

$\sf{S_{\alpha +433}-S_{\alpha }=404040}$

$\implies\sf{(\alpha +433)^2-\alpha ^2=404040}$

$\implies\sf{\alpha ^2+866\alpha +433^2-\alpha ^2=404040}$

$\implies\sf{866\alpha =404040-433^2}$

$\implies\sf{(even)=(even)-(odd)}$

This equation has no solution because the LHS and RHS never equal.

Hence, such a series never exists.