Geometric Sequence

Note that  the sequence is

1000 , 1000(-1/3)  , 1000(-1/3)^2 , 1000(-1/3)^3   ........

We only need to  consider the terms that are positive  (these will be the odd terms)

So

1000 (1/3)^(n -1)  >  1         where n is  the nth term

(1/3)^(n -1)  >  1/1000    take the log of  both sides

log (1/3)^(n -1) >  log (1/1000)       and we can write

(n -1) log (1/3) > log (.001)      

Divide  both sides by log (1/3)....since this is  negative, we need to reverse the inequality sign

n - 1   <  log (.001) / log (1/3)

n <    1 +  log (.001) / log (1/3)   ≈  7.28

Note  that  the 7th term    =  1.37

Note that the 9th term = .152

So  only  terms   1 , 3 , 5  and 7   are >  1

In a geometric sequence, the ratio between successive terms is constant. In this case, the common ratio is -1/3, which means each term is -1/3 times the term before it.

For the sequence to be greater than 1, the terms need to be positive. Since the first term (1000) is positive and the common ratio is negative, the sequence will alternate between positive and negative terms.

Since there are 400 terms, half (200) will be positive and the other half negative. However, the question asks for terms greater than 1, not just positive terms.

Finding the first term greater than 1:

We can analyze the sequence using the formula for the nth term of a geometric sequence:

tn = a * r^(n-1)

tn: nth term

a: first term (1000)

r: common ratio (-1/3)

n: term number

We want to find the value of n (n = k) where tk > 1.

Since the terms alternate in sign, we need to consider the even terms (positive terms). Let's rewrite the formula for even terms (n = 2k):

t(2k) = 1000 * (-1/3)^(2k-1)

For tk to be greater than 1, we need the exponent of -1/3 to be negative enough to make the result positive.

The smallest even term that satisfies this condition is t4 (fourth term), where k = 2.

t4 = 1000 * (-1/3)^(2(2)-1) = 1000 * (-1/9) = -100/9 < 1

t6 (sixth term) = 1000 * (-1/3)^(2(3)-1) = 1000 * (1/27) = 100/27 > 1

Therefore, the first term greater than 1 is the sixth term (n = 6).

Counting terms greater than 1:

Since the sequence alternates in sign, every two terms after the sixth term will have one term greater than 1. With 400 terms, there are 200 even terms (positive or negative).

Out of these 200 terms, the first 4 (t2, t4, t6, and t8) won't be greater than 1. The remaining 196 terms will have 98 terms greater than 1 (every other term).

So, there are 98 terms greater than 1 in the sequence.