# Question Video: Solving Word Problems Involving Arithmetic Series Mathematics

The house numbers on one side of a street are 1, 2, 3, 4, ..., 49. Find the house number where the sum of the numbers before it is equal to the sum of the numbers after it.

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### Video Transcript

The house numbers on one side of a street are one, two, three, four, and so on all the way up to 49. Find the house number where the sum of the numbers before it is equal to the sum of the numbers after it.

The counting or natural numbers from one to 49 form an arithmetic sequence. The first term of the sequence 𝑎 is equal to one. The common difference 𝑑 is also equal to one as the numbers are increasing by one. We know that the sum of 𝑛 terms in an arithmetic sequence is equal to 𝑛 over two multiplied by two 𝑎 plus 𝑛 minus one multiplied by 𝑑. We know in total that there will be 49 terms as there are 49 houses. However, we are looking for the house number where the sum of the numbers before it is equal to the sum of the numbers after it.

We can therefore let 𝑛 be the house number that we want. The sum of the houses before, written 𝑠 sub 𝑛 minus one, is therefore equal to 𝑛 minus one over two multiplied by two plus 𝑛 minus two. This is because two 𝑎 is equal to two multiplied by one, which is two. Our 𝑛-value from the general formula is now 𝑛 minus one. This means that we have 𝑛 minus two multiplied by the common difference one, which gives us 𝑛 minus two. The expression inside the parentheses simplifies to 𝑛 as two minus two is zero. This means that the sum of the first 𝑛-minus-one terms is 𝑛 over two multiplied by 𝑛 minus one. This is the sum of the house numbers before our house.

The sum of the house numbers after our house will be equal to the sum of all 49 house numbers minus 𝑠 sub 𝑛. The sum of all the house numbers is equal to 49 over two multiplied by two plus 48. This in turn simplifies to 49 over two multiplied by 50. 49 multiplied by 50 is 2450. And dividing this by two or halving it gives us 1225. The sum of the house numbers up to and including our house, 𝑠 sub 𝑛, is equal to 𝑛 over two multiplied by two plus 𝑛 minus one. The parentheses here simplifies to 𝑛 plus one as two minus one is equal to one. We can therefore conclude that the sum of the house numbers after our house is 1225 minus 𝑛 over two multiplied by 𝑛 plus one.

We need these two expressions to be equal. This means that 𝑛 over two multiplied by 𝑛 minus one is equal to 1225 minus 𝑛 over two multiplied by 𝑛 plus one. Multiplying both sides of our equation by two and distributing our parentheses gives us 𝑛 squared minus 𝑛 is equal to 2450 minus 𝑛 squared minus 𝑛. We can add 𝑛 to both sides of this equation. We can then add 𝑛 squared to both sides such that two 𝑛 squared is equal to 2450. Dividing by two gives us 𝑛 squared is equal to 1225. We can then square root both sides, and since 𝑛 must be positive, 𝑛 is equal to 35.

The sum of the house numbers before 35 is equal to the sum of the house numbers after it. We could check this by substituting 𝑛 equals 35 into our expressions for the houses before and houses after. In both cases, we get a total of 595. The sum of the integers from one to 34 is 595, and the sum of the integers from 36 to 49 is also 595. Therefore, the correct answer is 35.