Question

fill in the geometric means: 14.
x | 1 | 2 | 3 | 4 | 5
y | 6 | | | | 96
15.

Explanation:

Step1: Identify the geometric sequence

We have a geometric sequence where the first term \( a_1 = 6 \) (when \( x = 1 \)) and the fifth term \( a_5=96 \) (when \( x = 5 \)). The formula for the \( n \)-th term of a geometric sequence is \( a_n=a_1r^{n - 1} \), where \( r \) is the common ratio.

For \( n = 5 \), we have \( a_5=a_1r^{5 - 1}\), so \( 96=6\times r^{4} \).

Step2: Solve for the common ratio \( r \)

Divide both sides of the equation \( 96 = 6r^{4} \) by 6:
\( \frac{96}{6}=r^{4} \)
\( 16=r^{4} \)
Taking the fourth root of both sides, we get \( r=\pm2 \). We can consider the positive ratio first (we can check the negative later, but usually, for such problems, positive ratio is expected unless stated otherwise). So \( r = 2 \) (we can also check \( r=- 2 \), but let's proceed with \( r = 2 \) first).

Step3: Find the second term (\( x = 2 \))

The second term \( a_2=a_1r \). Substitute \( a_1 = 6 \) and \( r = 2 \):
\( a_2=6\times2 = 12 \)

Step4: Find the third term (\( x = 3 \))

The third term \( a_3=a_2r \). Substitute \( a_2 = 12 \) and \( r = 2 \):
\( a_3=12\times2=24 \)

Step5: Find the fourth term (\( x = 4 \))

The fourth term \( a_4=a_3r \). Substitute \( a_3 = 24 \) and \( r = 2 \):
\( a_4=24\times2 = 48 \)

Let's check with \( r=-2 \):
\( a_2=6\times(-2)=- 12 \), \( a_3=-12\times(-2) = 24 \), \( a_4=24\times(-2)=-48 \), \( a_5=-48\times(-2)=96 \). Both \( r = 2 \) and \( r=-2 \) work, but since the first term is positive and the fifth term is positive, both positive and negative ratios are possible. But usually, in such problems, positive ratio is more common. So the terms are 12, 24, 48 (or - 12, 24, - 48).

Answer:

For \( r = 2 \): The values of \( y \) when \( x = 2,3,4 \) are 12, 24, 48 respectively.

For \( r=-2 \): The values of \( y \) when \( x = 2,3,4 \) are - 12, 24, - 48 respectively. (But the most probable is 12, 24, 48)