Among the natural numbers 1 to 49, find the sum of all preceeding numbers and the sum of all he succeeding numbers are equal .
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let x be the number from 1 to 49 such that sum of its preceding is equal to the sum of numbers succeeding.
Now According to arithmetic progression
1+2+3+.....+(x-1)=(x+1)+(x+2)+(x+3)+.....+49
=> 1+2+3+.....+(x-1)=[1+2+3+.....+49]-(1+2+3+....+x)
=>x(x-1)upon2 =49×50upon2 - x(x-1)upon2
After solving above equation we get
2x² = 49×50
x² = 49×25
x² = 7² × 5²
x = 35
Hence, 35 is the required number between 1 to 49.
Now According to arithmetic progression
1+2+3+.....+(x-1)=(x+1)+(x+2)+(x+3)+.....+49
=> 1+2+3+.....+(x-1)=[1+2+3+.....+49]-(1+2+3+....+x)
=>x(x-1)upon2 =49×50upon2 - x(x-1)upon2
After solving above equation we get
2x² = 49×50
x² = 49×25
x² = 7² × 5²
x = 35
Hence, 35 is the required number between 1 to 49.
harshjari6251:
show me full method plz I can't understand properly
ok
To understand this question ; Lets say that from 1 to 49 there is a number x in anywhere from 1 to 49 like => 1,2,3,4,......(x-1),x,(x+1),...........47,48,49
In the question it is given that sum of the numbers who come before x like 1+2+3+.....+(x-1) is equal to sum of the numbers who come after x like (x+1)+......+47+48+49
Now just simply use the formula given in chapter A.P. which is about sum of numbers in an A.P.
first we will apply the formula on 1+2+3+....+(x-1) after that will apply it on (x+1)+......+47+48+49 and we will equate the values of both equations
If you understood the logic of this question then please help me to have a brainlist.
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