How i solve this question of arithmetic progressions of exercise 5.4 of class 10th math, how i solve this question in easy way The houses of a row are numbered consecutively from 1 to 49. Show that there is 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. Find this value of x. [Hint :Sx – 1 = S49 – Sx ]

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# The houses of a row are numbered consecutively from 1 to 49. Show that there is 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. Find this value of x. [Hint :Sx – 1 = S49 – Sx ] Q.4

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Given,

Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First term, a = 1

Common difference, d=1

Let us say the number of x

^{th}houses can be represented as;Sum of nth term of AP = n/2[2a+(n-1)d]

Sum of number of houses beyond x house = S

_{x-1}= (x-1)/2[2.1+(x-1-1)1]

= (x-1)/2 [2+x-2]

= x(x-1)/2 ………………………………………(i)

By the given condition, we can write,

S

_{49 }– S_{x}= {49/2[2.1+(49-1)1]}–{x/2[2.1+(x-1)1]}= 25(49) – x(x + 1)/2 ………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore,

x(x-1)/2 = 25(49) – x(x-1)/2

x = ±35

As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.