The sum of several consecutive natùral is 2012 . how many numbers were in that sum ?
Answers
Answer:
8
Step-by-step explanation:
The 8 consecutive numbers are 248,249,250,251,252,253,254,255.
The Sum of 8 consecutive natural numbers is 2012 and numbers are 248 , 249 , 250 , 251 , 252 , 253 , 254 , 255
Assume that there are n consecutive natural numbers whose sum is 2012 and first number is a
Sₙ = (n/2)(2a + (n - 1)d)
d = 1 as consecutive natural numbers
2012 = (n/2)(2a + (n - 1)1)
4024 = n ( 2a + n - 1)
Factors of 4024 are
1 , 2 , 4 , 8 , 503 , 1006 , 2012 , 4024
Hence Maximum possible value of n can be 8
as for n ≥ 503 value of a will be negative but a is natural number
if n = 8
Then 2a + n - 1 = 4024/8
=> 2a + 8 - 1 = 503
=> 2a = 496
=> a= 248
Numbers are 248 , 249 , 250 , 251 , 252 , 253 , 254 , 255
if n = 4
Then 2a + 4 - 1 = 4024/4
=> 2a + 3= 1006
=> 2a = 1003
=> a is not a natural number
if n = 2
Then 2a + 2 - 1 = 4024/2
=> 2a + 1= 2012
=> 2a =2011
=> a is not a natural number
n = 1 is not several numbers
Hence only possible solution is 8
Numbers are 248 , 249 , 250 , 251 , 252 , 253 , 254 , 255
The Sum of 8 consecutive natural numbers is 2012