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Q.

The houses of a row are numbered consecutively from 1 to 49. Find the 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. What will be the value of x?

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a

29

b

35

c

42

d

37

answer is B.

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Detailed Solution

It is given the houses of a row are numbered consecutively from 1 to 49 and sum of the numbers of the houses preceding the house numbered x = sum of the numbers of the house following it.
In an arithmetic progression let the first term be ‘a’ and common difference be ‘d’.
The

n^{th}

term of an AP is =

a + (n - 1) d

The sum of first n terms of an AP =

\frac{n}{2} (2 a + (n - 1) d)

According to the information given in the question, a = 1; d = 1
Number of houses preceding x = x -1
Sum of the number of houses preceding x =

\frac{(x - 1)}{2} (2 \times 1 + (x - 1 - 1) \times 1) = \frac{x (x - 1)}{2}

Number of houses succeeding x = 49 – x
Sum of the number of houses succeeding x,

\Rightarrow \frac{(49 - x)}{2} (2 \times (x + 1) + (49 - x - 1) \times 1) = \frac{(49 - x) (50 + x)}{2}

Equating the two equations according to the condition given we get,

\Rightarrow \frac{x (x - 1)}{2} = \frac{(49 - x) (50 + x)}{2}

{\Rightarrow x}^{2} - x = 2450 + 49 x - 50 x - x^{2}

\Rightarrow 2 x^{2} = 2450

{\Rightarrow x}^{2} = 1225

\Rightarrow x = \pm 35

Since x cannot be negative, we get x =35.
Hence option (2) is correct.

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