in sequence 1, the first term is 2 and every term after the first is 2 greater than the previous term....?

Question

In sequence 2, the first term is 1/64 and every term after the first is 2 times the previous term. What is the leaswt positive integer n for which the nth term of sequence 2 is greater than the nth term of sequence 1?



can someone please help me with this
and explain how you do the problem
thanx =]

Answer 1

The first is an arithmetic progression, whose nth term is 2n(even numbers sequence).
The second is a geometric progression whose nth term in ( 2^(n-1) ) / 64
Thus 2^(n-1) ) / 64 > 2n
=> 2^(n-8) > n
Thus the first n for which this inequality is true is n = 12.

Answer 2

One can also just compute the terms:

n s1 s2
1 2 1/64
2 4 1/32
3 6 1/16
4 8 1/8
5 10 1/4
6 12 1/2
7 14 1
8 16 2
9 18 4
10 20 8
11 22 16
12 24 32 ***

This is no more or less work, no more or less elegant than reducing things to 2n < 2^(n-1)/64 [which reduces further to n < 2^(n-8)], but perhaps more straightforward.

Answer 3

Usually the way to do problems like this is just actually write out some terms, as per the second answer you got.