find the 12th term of the arithmetic progression 1/2, 1/6, 1/10
1/10 = 3/30
1/6 = 5/30
---------- subtract
d = -2/30 = -1/15
check it first two points
d = 5/30 - 15/30 = -10/30 = -1/3
This is not an arithmetic progression, typo.
halfway between 1/2 and 1/10
is
(.5 + .1)/2 = .3 = 3/10 not 1/6
However, if we consider the sequence as
1/2 * 1/1, 1/2 * 1/3, 1/2 * 1/5
Tn = (1/2) * 1/(2n-1)
T12 = (1/2)(1/23) = 1/46
There are other rules which would produce the given terms, but this is the easiest for me to see.
To find the 12th term of an arithmetic progression, we need to know the first term and the common difference.
In this case, the first term is 1/2 and the second term is 1/6. To find the common difference, we subtract the second term from the first term.
common difference = 1/6 - 1/2 = (1/6) - (3/6) = -2/6 = -1/3
Now that we have the first term and the common difference, we can use the formula for the nth term of an arithmetic progression:
nth term = first term + (n-1) * common difference
Substituting the values we have:
12th term = 1/2 + (12-1) * (-1/3)
= 1/2 + 11*(-1/3)
= 1/2 - 11/3
= (3/6) - (22/6)
= -19/6
Therefore, the 12th term of the arithmetic progression 1/2, 1/6, 1/10 is -19/6.