Question

which term of the sequence 2,√2,4 is 128

Answer 1

n=l-a/d+1
=128-2/root2-4+1

Answer 2

Given series is 2,22–√,4,.........2,22,4,.........

This series is G.P. since the ratio of any two successive terms is same.

first term=a=2=a=2 and common ratio r=22–√2r=222=2–√=2

We know that nth term =tn=a.rn−1=tn=a.rn−1

Given that tn=128tn=128

⇒128=2.(2–√)n−1⇒128=2.(2)n−1

2–√=21/22=21/2 and 128=27128=27

⇒27=2(1+n−12)⇒27=2(1+n−12)

Since the base on either side is 2, power on both sides are equal.

⇒7=1+n−12=n+12⇒7=1+n−12=n+12

or n=13n=13

i.e.,13thi.e.,13th term of the sequence is 128