Question

Which term of the GP : 2, 2√2, 4 ..... is 128 ?

Answer 1

Hi ,

Let a and r are first term and common

ratio of a G.P

nth term = an = ar^n-1

Here ,

Given G.P is 2 , 2√2 , 4 .,,

a = 2 ,

r = a2/A1

r = 2√2/2

r = √2

Let n th term = 128

ar^n-1 = 128

2 × (√2)^ n - 1 = 128

2^( n - 1 )/2 = 128/2

2^( n - 1 )/2 = 64

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

( n - 1 )/2 = 6

[ Since i ) 64 = 2^6 ,



ii ) If a^m = a^n then m = n ]

n - 1 = 12

n = 12 + 1

n = 13


Therefore ,

13th term of given G.P is 128.

I hope this helps you.

: )

Answer 2

Answer:

r=2root2/2=root2

Step-by-step explanation:

so, 128=2*root2powern

64=root2power n

2to the power 6=2power1/2n

n=13

it will be helpful to you