Question

The houses of a row are numbered consecutively from 1 to 49. show that there is a value of x such that sum of the numbers of houses preceding house numbered x is equal to the sum of the nums of houses following it. Find the value of x​

Answer 1

Answer:

don't no sorry really ok

Answer 2

$\huge{\boxed{\mathtt{\red{ANSWER}}}}$

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)

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