Question

How many terms are there in the sequence of 1/128, 1/64, 1/32....., 32, 64?
(a) 13
(b) 14
(c) 15
(d) 16​

Answer 1

Answer:

C IS THE ANSWER OK BYEBYEBYE

Answer 2

Given:

1/128, 1/64, 1/32,.., 32, 64

To find:

The number of terms in the sequence

Solution:

The number of terms in the sequence is 14. (Option b)

We can find the number of terms in the following way-

We know that the given sequence of numbers is a geometric progression.

The first term of the G.P.=1/128

The last term of the G.P.=64

The common ratio=second term/first term

=(1/64)/(1/128)

=1/64×128/1

=128/64

=2

Let the number of terms in the G.P. be n.

We know that the nth term of a G.P.=$ar^{n-1}$

Here, the nth term is the last term of the G.P., a is the first term, and r is the common ratio.

On putting the values, we get

64=1/128×$2^{n-1}$

64×128=$2^{n-1}$

$2^{6}$×$2^{7}$=$2^{n-1}$

$2^{13}$ =$2^{n-1}$

Now, since the base is equal, powers will also be equal.

So, 13=n-1

13+1=n

n=14

Therefore, the number of terms in the sequence is 14.