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The houses of arrows are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the number of houses preceding the house number x is equal to the sum of the number of the houses following it. Find the value of x.

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Last updated date: 18th Jun 2024
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Answer
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Hint: In this question, we will first write the house number and find that the numbering is forming an AP whose common difference is 1. After this we will write the sum of house number before house x and sum of house numbers after house x and then equate them to get the equation.

Complete step-by-step answer:
In this question we are going to find the value of x.
Let there be a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it .
The house number is 1, 2 , 3 …………….x-1,x, x+1……….49.
So, we have a sum of numbers from 1 to x-1 = sum of numbers from x+1 to 49.
Now it is clear from the above data that house number will form an A.P whose
First term and the common difference is 1.
Now we know that the Sum of n terms of an AP is given as:
${{\text{S}}_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ , where
n = number of terms in AP.
a = first term of AP.
d= first term of AP.
Let ${S_1}$ be the sum of numbers from 1 to x-1.
${S_1}$= 1+2+3+........+x-1.
Here a =1, d = 1 and n= x-1.
Putting these values in formula of sum, we get:
${{\text{S}}_1} = \dfrac{{x - 1}}{2}\left( {2 \times 1 + \left( {x - 1 - 1} \right)1} \right) = \dfrac{{x(x - 1)}}{2}$
Let ${S_2}$ be the sum of numbers from x+1 to 49.
${S_2}$= (x+1) + (x+2) +........+48+49.
Here a =x+1, d = 1 and n= 49-(x+1)+1=49-x.
Putting these values in formula of sum, we get:
${{\text{S}}_2} = \dfrac{{49 - x}}{2}\left( {2 \times (x + 1) + \left( {49 - x - 1} \right)1} \right) = \dfrac{{(49 - x)(x + 50)}}{2}$
Now according to question:
${S_1}$=${S_2}$
Putting the values of ${S_1}$and ${S_2}$in above equation, we get:
$\dfrac{{x(x - 1)}}{2}$ =\[\dfrac{{(49 - x)(x + 50)}}{2}\]
On further solving , we get:
$ \Rightarrow {x^2} - x = 49x + 2450 - {x^2} - 50x$
$ \Rightarrow {x^2} - x = 2450 - {x^2} - x$
Taking all terms on LHS, we get:
$ \Rightarrow 2{x^2} = 2450$
$ \Rightarrow {x^2} = \dfrac{{2450}}{2} = 1225$
On taking square root on both sides, we get:
$x = \sqrt {1225} = 35$
Hence the value of x which is satisfying our given question is 35.

Note: In this type of question, you should follow the information given the question and then use the result of the AP or GP according to the type of sequence. You should remember the formula for nth term of AP which is given as:
${T_n}$ =a+(n-1)d, where, a is first term , n is value of nth term and ‘d’ is a common difference.