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

There are 7 terms in the geometric progression 2, 4, 8, ...., 128.

To find the number of terms in a geometric progression, we can use the formula:

n = log(base r) (a_n / a_1) + 1

where:
n = number of terms
r = common ratio
a_n = nth term
a_1 = first term

In this case, the common ratio is 4/2 = 2, the first term is 2, and the nth term is 128. Plugging these values into the formula:

n = log(base 2) (128 / 2) + 1
n = log(base 2) 64 + 1
n = 6 + 1
n = 7

Therefore, there are 7 terms in the geometric progression 2, 4, 8, ...., 128.

The first term (a) is 2 and the common ratio (r) is 4/2 = 2.
The last term (l) is 128.

The formula for the nth term of a geometric progression is given by:

nth term (Tn) = a - r^(n-1)

Substituting the given values into the formula, we get:

128 = 2 - 2^(n-1)

Dividing both sides by 2, we get:

64 = 2^(n-1)

Taking the logarithm of both sides to base 2, we get:

log2(64) = log2(2^(n-1))

Simplify both sides:

6 = n-1

Adding 1 to both sides, we get:

n = 7

Therefore, there are 7 terms in the geometric progression 2, 4, 8, ...., 128.