1.Given a geometric sequence with a_1=6 and r=2/3, write an explicit formula for a_n, the nth term of the sequence.

Plug and chug:

http://www.mathsisfun.com/algebra/sequences-sums-geometric.html

ok thanks i get #2 but i still don't understand 1.

an = a1 r^(n-1) = 6 * (2/3)^(n-1)

= 6 * 2(n-1) /3^(n-1)

1. To find the explicit formula for the nth term, we can use the formula for a geometric sequence:

aₙ = a₁ * r^(n-1),

where aₙ is the nth term, a₁ is the first term, r is the common ratio, and n is the position of the term.

In this case, a₁ = 6 and r = 2/3. Substituting these values into the formula, we get:

aₙ = 6 * (2/3)^(n-1).

Therefore, the explicit formula for the nth term of the sequence is aₙ = 6 * (2/3)^(n-1).

2. We are given a₄ = 4 and a₅ = 7 for a geometric sequence. From the formula for a geometric sequence, we have:

aₙ = a₁ * r^(n-1),

where aₙ is the nth term, a₁ is the first term, r is the common ratio, and n is the position of the term.

We can use the given values to form two equations:

a₄ = a₁ * r^(4-1) = a₁ * r³ = 4 (Equation 1)
a₅ = a₁ * r^(5-1) = a₁ * r⁴ = 7 (Equation 2)

To solve these equations, we can divide Equation 2 by Equation 1:

(a₁ * r⁴) / (a₁ * r³) = 7 / 4
r = 7/4

Now, we can substitute the value of r back into either Equation 1 or Equation 2 to find the value of a, the first term. Let's use Equation 1:

a₁ * (7/4)³ = 4

Simplifying the equation:

a₁ * (343/64) = 4
a₁ = (4 * 64) / 343
a₁ = 256 / 343

Therefore, a = a₁ = 256/343.