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Grazes
In a certain street, there are more than fifty but less than five hundred houses in a row, numbered from 1, 2, 3 etc. consecutively. There is a house in the street, the sum of all the house numbers on the left side of which is equal to the sum of all house numbers on its right side. Find the number of this house. Also find all solutions to this problem, without the above 50 and 500 constraint. Try to find a method to solve this without guess and check.
use an arithmetic sequence here. recall the sum of the first n terms of an arithmetic sequence is given by:\[S_n=\frac{n}{2}(2a_1+(n-1)d)\]where \(a_1\) is the first term, \(n\) is the number of terms and \(d\) is the common difference between each term. In your case: \(a_1=1\) and \(d=1\) which means we can simplify the sum to:\[S_n=\frac{n}{2}(2+(n-1))=\frac{n(n+1)}{2}\tag{a}\] Now, let's say that the i'th house satisfies the conditions in your question, then that would imply:\[S_{i-1}=S_n-S_i\tag{b}\]because \(S_{i-1}\) is the sum of the house numbers to the left of the i'th house, and \(S_n-S_i\) is the sum of the house numbers to the right of the i'th house. If we now substitute the formula in (a) into the relation in (b) we get:\[\frac{(i-1)(i-1+1)}{2}=\frac{n(n+1)}{2}-\frac{i(i+1)}{2}\]multiplying both sides by 2 we get:\[(i-1)(i)=n(n+1)-i(i+1)\]therefore:\[i(i-1)+i(i+1)=n(n+1)\]which simplifies to:\[2i^2=n(n+1)\tag{c}\]so you now need to find integer solutions to equation (c) for \(i\) and \(n\).
if there are 8 houses then the number the house is 6 which the sum of all the house numbers on the left side of which is equal to the sum of all house numbers on its right side. AND THERE is NO SOLUTION IF THERE ARE fifty but less than five hundred houses in A ROW