Video: CBSE Class X • Pack 3 • 2016 • Question 27

CBSE Class X • Pack 3 • 2016 • Question 27

05:02

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.

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