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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Answers
Answer:
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 xth 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 = Sx-1
= (x-1)/2[2.1+(x-1-1)1]
= (x-1)/2 [2+x-2]
=__________________(i)
By the given condition, we can write,
S49 – Sx = {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,
As we know, the number of house cannot be an a negative number. Hence, the value of x is 35.
Answer:
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