Question

A geometric series has a constant ratio of 21​ and a sum to infinity of 6 . (2) 4.1 Calculate the first term of the series. 4.2 Calculate the 8th term of the series. (2) 4.3 Given: ∑k=1n​3(2)1−k=5,8125 Calculate the value of n. (4) 4.4 If ∑k=120​3(2)1−k=p, write down ∑k=120​24(2)−k in terms of p. (3) [11]

Answer 1

The first term of the geometric series is -120. The 8th term of the series is -75295360. The value of n is approximately 8. The series ∑k=120​24(2)−k is equal to p.

Step-by-step explanation:

To find the first term of the geometric series, we can use the formula for the sum to infinity of a geometric series, which is S = a / (1 - r), where S is the sum to infinity, a is the first term, and r is the common ratio. In this case, we are given that the sum to infinity is 6 and the common ratio is 21. Plugging these values into the formula, we get 6 = a / (1 - 21). Solving for a, we find that the first term of the series is -120.

To find the 8th term of the series, we can use the formula for the nth term of a geometric series, which is an = a * r^(n-1), where an is the nth term, a is the first term, r is the common ratio, and n is the term number. In this case, we are given that the first term is -120 and the common ratio is 21. Plugging these values into the formula, we get a8 = -120 * 21^(8-1). Evaluating this expression, we find that the 8th term of the series is -75295360.

To find the value of n in the given series, we can use the formula for the sum of a geometric series, which is Sn = a * (1 - r^n) / (1 - r), where Sn is the sum of the first n terms, a is the first term, r is the common ratio, and n is the number of terms. In this case, we are given that the sum is 58125 and the common ratio is 2. Plugging these values into the formula, we get 58125 = a * (1 - 2^n) / (1 - 2). Simplifying the equation, we get 116250 = a * (1 - 2^n). We can solve this equation by trial and error or by rearranging it to find an equivalent equation. In this case, we can rearrange the equation to get 2^n = 1 - 58125 / a. Since a is unknown, we can substitute the value of a we found in part 4.1 (-120) and solve for n. Plugging in the values, we get 2^n = 1 - 58125 / (-120). Simplifying the expression, we get 2^n = 1 + 484.375. Taking the logarithm of both sides, we get n * log(2) = log(485.375). Dividing both sides by log(2), we find that n is approximately 8.

To write down the series ∑k=120​24(2)−k in terms of p, we need to determine the value of p. We are given that ∑k=120​3(2)1−k = p. We can rewrite this as ∑k=120​2(2)−k, which is equivalent to ∑k=120​24(2)−k. Therefore, the series ∑k=120​24(2)−k is equal to p.