Question

The houses of a row are numbered consecutively from 1 to 49. Show that there is value of x such that the sum 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.

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)

Answer 2

The total number of houses preceding the house numbered x = (x-1)

and the number of housed following the house numbered x = 49-x

The house number following the house number x = x+1

According to quesiton

Sx-1 = S4 9-x

For the house preceding house number x:a = 1, d = 1, n = x-1

For the house following house number x:a = x+1, d = 1, last term, l = 49

So,

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

[(x-1)/2]92+x-2) = [(49-x)/2](50+x)

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

x2 - x = 2450 + 49x - 50x - x2

x2 - x = 2450 - x - x2

x2 - x + x + x2 = 2450

2x2 = 2450

x2 = 2450/2 = 1225

x = sqrt(1225)

x = 35