### Video Transcript

A row of houses are numbered
consecutively from one to 49. Show that there exists a value of
𝑥 such that the sum of the numbers of the houses preceding the house numbered 𝑥 is
equal to the sum of the numbers of the houses following it.

It may not look like it, but this
question is testing our understanding of arithmetic progressions. Remember an arithmetic progression
is a sequence of numbers such that the difference between any two consecutive terms
is constant.

Since the houses are numbered
consecutively, the house numbers can be written as an arithmetic progression with a
first term of one and a common difference of one. The house number 𝑥 lies somewhere
in between the numbers one and 49. Remember the value of 𝑥 needs to
be an integer. We know that for two reasons. Firstly, we’re told that the row of
houses are numbered consecutively that implies one after the other, but also they
are house numbers. It makes no sense to have a
non-integer house number like 21.7 or a third.

We are going to be calculating the
sum of a given number of terms. So we need to recall the formula
for the sum of the first 𝑛 terms of an arithmetic progression. The sum of the first 𝑛 terms of an
arithmetic progression with a first term 𝑎 and a common difference 𝑑 is given by
𝑛 over two multiplied by two 𝑎 plus 𝑛 minus one multiplied by 𝑑.

Remember though we said that our
sequence has a first term of one and a common difference of one. So we can replace 𝑎 and 𝑑 with
one in our formula. Two multiplied by one is two and 𝑛
minus one multiplied by one is 𝑛 minus one. Two minus one then simplifies
further to one. So the formula for the first 𝑛
terms for our sequence is given by 𝑛 over two multiplied by one plus 𝑛.

We’re looking to find the house
number 𝑥 for which the sum of the numbers of the houses preceding this one — that’s
house numbers one to 𝑥 minus one — is equal to the sum of the numbers of the houses
following it — that’s house numbers 𝑥 plus one all the way up to 49.

We can begin by finding the sum of
the numbers of the houses from one to 𝑥 minus one by substituting 𝑥 minus one for
𝑛 in the formula for the sum of the first 𝑛 terms. That becomes 𝑥 minus one over two
multiplied by one plus 𝑥 minus one. One minus one is zero. So our formula for the sum of the
first 𝑥 minus one terms is 𝑥 multiplied by 𝑥 minus one all over two.

Next, to find the sum of the terms
from 𝑥 plus one to 49, we need to subtract the sum of the first 𝑥 terms from the
sum of all 49. The sum of the first 49 terms is 49
over two multiplied by one plus 49. 49 plus one is 50. So this simplifies to 49 multiplied
by 50 over two. Now, we could evaluate this
expression, but let’s leave it in this format for now. The sum of the first 𝑥 terms is
simply 𝑥 over two multiplied by one plus 𝑥. This is the same as 𝑥 multiplied
by one plus 𝑥 all over two. The difference between these is 49
multiplied by 50 over two minus 𝑥 multiplied by one plus 𝑥 again over two.

Since we’re trying to find the
value of 𝑥 for which the sum of the first 𝑥 minus one terms is equal to the sum of
the terms from 𝑥 plus one to 49, we can equate these two expressions. That gives us a single equation in
terms of 𝑥. We can multiply through by two. And then, we should expand each of
the brackets by multiplying every term inside the bracket by the number on the
outside. That gives us 𝑥 squared minus 𝑥
is equal to 49 multiplied by 50 minus 𝑥 squared minus 𝑥.

We can add an 𝑥 to both sides. And next, we should add 𝑥 squared
to both sides. That gives us that two 𝑥 squared
is equal to 49 multiplied by 50. Next, we should divide through by
two. 49 multiplied by 50 all divided by
two is the same as 49 multiplied by 50 divided by two and 50 divided by two is
25.

To solve for 𝑥, we’ll find the
square root of both sides. That gives us 𝑥 is equal to plus
or minus the square root of 49 multiplied by 25. Our house numbers are positive
though. So we can disregard the negative
version. We can also write this as the
square root of 49 multiplied by the square root of 25. The square root of 49 is seven and
the square root of 25 is five. That gives us 𝑥 is equal to
35.

Since this is an integer value, we
know that there does indeed exist a house number which satisfies the criteria
given. That’s house number 35.