Please! If the sixth term in a geometric sequence is 1/625, and the common ratio is 1/5, find the explicit formula of the sequence.

Question 21 options:

an=(15)n−1for n ≥ 1

an=(5)⋅(15)n−1for n ≥ 2

an=(1)⋅(15)n−1for n ≥ 2

an=(5)⋅(15)n−1for n ≥ 1

aₙ = 5*(1/5)^(n - 1)

Step-by-step explanation:

The recursive relation for a geometric sequence is:

aₙ = aₙ₋₁*R

where R is the common ratio.

We could write this in a general form as:

aₙ = a₁*(R)^(n - 1)

where a₁ is the first term of the sequence.

In this case we know that:

R = 1/5

And that:

a₆ = 1/625

Then we can replace these values in the above relation to get:

a₆ = (1/625) = a₁*(1/5)^(6 - 1) = a₁*(1/5^5) = a₁*(1/3125)

1/625 = a1*(1/3125)

(3125/625) = a1 = 5

Then the formula will be:

aₙ = 5*(1/5)^(n - 1)

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo

41*67/2=1373.5

step-by-step explanation:

i just used a calculator