What is the value of a1 for a geometric sequence with a4=40 and a6=160?

5

Step-by-step explanation:

The nth term of a geometric series is:

a_n = a₁ (r)^(n-1)

where a₁ is the first term and r is the common ratio.

Here, we have:

40 = a₁ (r)^(4-1)

160 = a₁ (r)^(6-1)

40 = a₁ (r)^3

160 = a₁ (r)^5

If we divide the two equations:

4 = r^2

r = 2

Now substitute into either equation to find a₁:

40 = a₁ (2)^3

40 = 8 a₁

a₁ = 5

(-infinity,infinity)

no it isn’t   i believe it is

step-by-step explanation: