A geometric sequence has a constant ratio between consecutive terms. In this case, 6/3 = 2, 12/6 = 2, and 24/12 = 2, confirming that the sequence is geometric.

To find the common ratio, divide any term by its previous term. Here, 15 ÷ 5 equals 3, which is consistent through the sequence.

A geometric sequence is formed by multiplying each term by a constant factor. In this case, every term is half of its predecessor, which makes the sequence geometric.

In a geometric sequence, the missing term is the geometric mean of the known terms. Since the geometric mean of 2 and 8 is √(2×8) = 4, the missing term is 4.

The sequence starts with 3 as the first term and each term is multiplied by 3. Counting the terms: 3 (1st), 9 (2nd), 27 (3rd), and 81 (4th) shows that 81 is the fourth term.

Dividing the second term by the first gives -1/3 and subsequent divisions yield the same ratio. This constant multiplication factor proves the sequence is geometric.

The common ratio is determined by dividing any term by its preceding term. Here, 6 divided by -2 equals -3, a result that continues consistently in the sequence.

Using the formula a(n) = a₝ × r^(n-1) with a₝ = 4 and r = 2, the 6th term is calculated as 4 × 2^(5) = 128. This confirms the correct term in the sequence.

A geometric sequence requires a constant multiplication factor between terms. The sequence 5, 10, 15, 20 increases by addition instead, making it arithmetic rather than geometric.

The nth term of a geometric sequence is given by a(n) = a₝ × r^(n-1). Substituting a₝ = 11, r = 1/2, and n = 4, we get 11 × (1/2)³ = 11/8.

By adding the terms 3, 9, 27, and 81, the total sum is 3 + 9 + 27 + 81 = 120. This simple addition confirms the correct sum for the sequence.

Using the relation a₅ = a₂ × r³, we find r³ = 96/12 = 8. Taking the cube root of 8 results in r = 2.

The formula for the nth term of a geometric sequence is a(n) = a₝ × r^(n-1). Other formulas represent arithmetic sequences or are not standard for geometric progressions.

The common ratio for the sequence is -2. Thus, the 5th term is calculated as 1 × (-2)^(4) = 16, where the even exponent results in a positive term.

Dividing the sum equation 2 + 2r + 2r² = 14 by 2 gives 1 + r + r² = 7. Solving the quadratic equation r² + r - 6 = 0, the positive integer solution is r = 2.

For a convergent geometric series, the sum to infinity is calculated by a/(1 - r). Here, with a = 16 and r = 0.5, the sum is 16/(1 - 0.5) = 32.

Using the relation a₅ = a₂ × r³, we substitute the values to get 243 = 9 × r³. Solving for r³ gives 27 and thus r = 3, considering the positive constraint.

Using the formula a(n) = a₝ × r^(n-1), we set up the equation 5 × 2^(n-1) = 320. Dividing by 5 gives 2^(n-1) = 64, and recognizing that 64 = 2❶ leads to n - 1 = 6, or n = 7.

Using the sum formula Sₙ = a × (1 - r❿)/(1 - r) for n = 5 with r = 0.5, the equation becomes a × (1 - 0.5❵)/(0.5) = 31. This simplifies to a × (31/16) = 31, so solving for a gives 16.

Since a₂ = a₝ × r = 9 and a₄ = a₝ × r³ = 81, dividing the two gives r² = 9, so r = 3 (considering only the positive solution). Thus, the first term is a₝ = 9/3 = 3.