Question

how many terms are there in a sequence of 1/128,1/64,1/32,......,32,64?

Answer 1

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

    the .......is = 1/16,1/8,1/4,1/2,0.5,1,2,4,8,16,then 32,64    

Answer 2

Answer:

The number of terms in given sequence is 14

Step-by-step explanation:

Given sequence is $\frac{1}{128}$, $\frac{1}{64}$, $\frac{1}{32}$,... 32, 64

which is in GP or Geometric Progression

here we need to find number of terms in progression

let 64 is nth term in given sequence, then we can say that given GP has n terms

 ⇒common ratio (r) in Geometric progression  =  $\frac{T_{n} }{T_{n-1} }$  

 ⇒ r =  $\frac{\frac{1}{64} }{\frac{1}{128} }$  =  $\frac{128}{64}$ =  2 ,  

 ⇒ first term $a =$  1/128  

nth term in GP = Tn = $ar^{n-1}$  =    $\frac{1}{128} ( 2^{n-1})$  =  64

                                           ⇒   $2^{n-1}$  =  64 (128)

                                           ⇒  $2^{n-1} = 8192$

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

                                           ⇒   n-1 =13  

                                           ⇒   n = 14

the number of terms in given sequence  is 14