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Answer :
1) let n be first integer. Next consecutive integer is n+1
If their sum is 141 we can write:
n + (n+1) = n + n + 1 = 2n + 1 = 141
2) now this is bit tricky. First integer is:
2n-1
the reason for this is because 2n is always even number for any n. when we subtract 1 we always get odd number. Next consecutive odd number is:
2n-1 + 2 = 2n+1
when we sum these 2 numbers we get:
2n-1 + (2n+1) = 2n - 1 + 2n + 1 = 4n.
Now the important part. 4n and 2n are both always even numbers so it doesnt matter if we write 2n or 4n or 6n. If you give yourself thought experiment you will see that sum of two consecutive odd integers is always even integer.
which means that we can write our formula in multiple ways and one of them is:
2n + 2.
2n is always even number and if we add 2 again we get even number.
answer is 2n+2
If their sum is 141 we can write:
n + (n+1) = n + n + 1 = 2n + 1 = 141
2) now this is bit tricky. First integer is:
2n-1
the reason for this is because 2n is always even number for any n. when we subtract 1 we always get odd number. Next consecutive odd number is:
2n-1 + 2 = 2n+1
when we sum these 2 numbers we get:
2n-1 + (2n+1) = 2n - 1 + 2n + 1 = 4n.
Now the important part. 4n and 2n are both always even numbers so it doesnt matter if we write 2n or 4n or 6n. If you give yourself thought experiment you will see that sum of two consecutive odd integers is always even integer.
which means that we can write our formula in multiple ways and one of them is:
2n + 2.
2n is always even number and if we add 2 again we get even number.
answer is 2n+2
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