Which term of the series 2187,729,243...is 1/9

Step-by-step explanation:

Givens

a = 2187

r = 1/3

L = 1/9

what is n

Formula

L = a*r^(n-1)

1/9 = 2187 * (1/3)^(n-1)

1/(9 * 2187) = (1/3)^(n - 1)

1/19683 = (1/3)^(n - 1)

You need to break 19683 down into its prime factors.

19683 = 3* 729

19683 = 3 * 3 * 243

19683 = 3 * 3 * 3 * 81

81 = 3^4 so there are 4 more threes in the breakdown.

19683 = 3 * 3 * 3 * 3 * 3 * 3 * 3

1/19683 = 1/(3)^(7-1)

1/9 is the 6th term in the series.

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.

only 10/60 isn't equivalent

hope this : )