The given sequence is a geometric progression (GP) where each term is twice the previous term.

The general formula for the nth term of a GP is ar^(n-1), where a is the first term, r is the common ratio, and n is the term number.

In this case, a = log2 4 = 2 (since 2^2 = 4), r = 2 (since each term is twice the previous term), and the nth term = 21.

So, we have 2 * 2^(n-1) = 21.

Solving for n, we get n = log2(21/2) + 1 = log2 10.5 + 1.

Therefore, n/4 = (log2 10.5 + 1) / 4.

To find the exact value, you would need to use a calculator.

Question

The given sequence is a geometric progression (GP) where each term is twice the previous term.

The general formula for the nth term of a GP is ar^(n-1), where a is the first term, r is the common ratio, and n is the term number.

In this case, a = log2 4 = 2 (since 2^2 = 4), r = 2 (since each term is twice the previous term), and the nth term = 21.

So, we have 2 * 2^(n-1) = 21.

Solving for n, we get n = log2(21/2) + 1 = log2 10.5 + 1.

Therefore, n/4 = (log2 10.5 + 1) / 4.

To find the exact value, you would need to use a calculator.

Knowee AI · Accepted Answer

The given sequence is a geometric progression (GP) where each term is twice the previous term.

The general formula for the nth term of a GP is ar^(n-1), where a is the first term, r is the common ratio, and n is the term number.

In this case, a = log2 4 = 2 (since 2^2 = 4), r = 2 (since each term is twice the previous term), and the nth term = 21.

So, we have 2 * 2^(n-1) = 21.

Solving for n, we get n = log2(21/2) + 1 = log2 10.5 + 1.

Therefore, n/4 = (log2 10.5 + 1) / 4.

To find the exact value, you would need to use a calculator.

log2 4 * log4 8 * log8 16 * ……………nth term = 21, what is the value of n/4?