In an Arithmetic Progression:
an = a1 + ( n - 1 ) d
where
a1 = the initial term
d = the common difference of successive members is d,
an = then the nth term
The even numbers on this sequence are members of the arithmetic progression:
a1 = 260 , an = 340 , d = 2
an = a1 + ( n - 1 ) d
340 = 260 + ( n - 1 ) ∙ 2
340 = 260 + 2 n - 2
340 = 258 + 2 n
Subtract 258 to both sides
82 = 2 n
Divide both sides by 2
41 = n
n = 41
The odd numbers on this sequence are members of the arithmetic progression:
a1 = 261 , an = 339 , d = 2
an = a1 + ( n - 1 ) d
339 = 261 + ( n - 1 ) ∙ 2
339 = 261 + 2 n - 2
339 = 259 + 2 n
Subtract 259 to both sides
80 = 2 n
Divide both sides by 2
40 = n
n = 40
There are 41 even numbers and 40 odd numbers in that sequence.
The number of even numbers between 260 and 340 is equal to the number of odd numbers between 260 and 340.Is it true , Prove
2 answers
If you include both end values, there cannot be the same number of evens and odds. There is an odd number of integers in the list, which cannot be evenly divided by 2.