Question

How many terms are there in the geometric progression 2, 4, 8...,128?

Answer 1

[AI-generated answer — Bot]

We can find the number of terms in a geometric progression by using the formula:

Number of terms = (logarithm of the last term over the first term) / (logarithm of the common ratio + 1)

In this case, the first term is 2, the last term is 128, and the common ratio is 4/2 = 2.

So,

Number of terms = (log 128 / log 2) / (log 2 + 1)

Number of terms = (7 / 1) / (1.301 + 1)

Number of terms = 7 / 2.301

Number of terms = 3.036

Therefore, there are approximately 3 terms in the geometric progression 2, 4, 8...,128.

Answer 2

hmmm

Answer 3

What a lot of gobble-de-goop by the bot !!

we have a GS, where a = 2, r = 2 and term(n) = 128
term(n) = a r^(n-1)
128 = 2(2^(n-1))
2^(n-1) = 64 = 2^6
n - 1 = 6
n = 7

There are 7 terms in the GS

All you had to do it extend the series a few more terms, so easy !!

Answer 4

[AI-generated answer — Explain Bot]

To find the number of terms in a geometric progression, you need to determine the common ratio and the first term. In this case, the common ratio between consecutive terms is 4 ÷ 2 = 2, and the first term is 2.

To find the number of terms, we can use the formula for the nth term of a geometric progression:

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 number of terms.

In this case, we have:

128 = 2 * 2^(n-1).

Dividing both sides of the equation by 2 gives:

64 = 2^(n-1).

We can rewrite the equation in exponential form:

2^6 = 2^(n-1).

Since the bases are equal, the exponents must be equal, so:

6 = n - 1.

Adding 1 to both sides yields:

n = 6 + 1 = 7.

Therefore, the geometric progression 2, 4, 8...,128 has 7 terms.