Question

The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value
of X such that sum of numbers of houses preceding the house numbered X is equal to sum
of the number of houses following X. Find the value of X.
​

Answer 1

Given 1,2,3,4,5,....49 consecutive numbers

x is the number such that

sum of preceding numbers of x = sum of following numbers of x

sum of ( 1,2,3,....x-1) = sum of [(x+1), (x+2) ,....48,49]

(x-1)/2[1+x-1] =(49-x)/2[x+1+49]

(x-1)x=(49-x)(x+50)

x²-x=49x+2450-x²-50x

x²-x =2450-x²-x

x²+x²-x+x=2450

2x²=2450

x²=2450/2

x²=1225

x=√1225

x=35

required number is x= 35

sum of (1,2,3.....34) = sum of (36,37,....49)

Hope it helps you

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Answer 2

Given :

1, 2, 3, ..... 49 consecutive numbers.

Sum of preceding numbers of x = Sum of following numbers of x.

i.e.

Sum of (1, 2, 3, .... x-1) = Sum of [(x+1), (x+2), (x+3), ..... 48, 49]

Find :

Value of x.

Solution :

Now,

$\Rightarrow\:S_1\: =\: 1\: + \:2 \:+\: 3 \:+ .... +\: (x \:-\: 1)$

$\Rightarrow\:S_1 \:= \:\bigg(\dfrac{x - 1}{2} \bigg)[2 \: \times \: 1 \: + \: (x - 1 - 1)1]$

$\Rightarrow\:S_1 \:= \:\bigg(\dfrac{x - 1}{2} \bigg)[2 \: + \: (x - 2)1]$

$\Rightarrow\:S_1 \:= \:\bigg(\dfrac{x - 1}{2} \bigg)(2 \: + \: x - 2)$

$\Rightarrow\:S_1 \:= \:\bigg(\dfrac{x - 1}{2} \bigg)(x)$

$\Rightarrow\:S_1 \:= \:x\bigg(\dfrac{x - 1}{2} \bigg)$

Now,

$\Rightarrow\:S_2 \:= \:\bigg(\dfrac{49 - x}{2} \bigg)[2 (x + 1 )\: + \: (49 - x - 1)1]$

$\Rightarrow\:S_2 \:= \:\bigg(\dfrac{49 - x}{2} \bigg)(2x \:+ \: 2 \: - \: x \:+ \:48)$

$\Rightarrow\:S_2 \:= \:\bigg(\dfrac{49 - x}{2} \bigg)(x\:+\:50)$

According to question,

$\bold{S_1\:=\:S_2}$

So,

$\Rightarrow\:x\bigg(\dfrac{x - 1}{2} \bigg)\:=\:\bigg(\dfrac{49 - x}{2} \bigg)(x\:+\:50)$

$\Rightarrow\:x(x\:-\:1)\:=\:(49\:-\:x)(x\:+\:50)$

$\Rightarrow\:x^2\:-\:x\:=\:49x\:+\:2450\:-\:x^2\:-\:50x$

$\Rightarrow\:x^2\:-\:x\:=\:-\:x^2\:-\:x\:+\:2450$

$\Rightarrow\:2x^2\:=\:2450$

$\Rightarrow\:x^2\:=\:1225$

$\Rightarrow\:x\:=\:35$

∴ Number is 35.

Sum of 1, 2, 3, .... (35 - 1) = Sum of [(35 + 1), (35 + 2) .... 48, 49)]

Sum of 1, 2, 3, .... 34 = Sum of (36, 37, .... 48, 49)