Question

Let three positive numbers a, b c are in geometric progression, such that a, b+8, c are in arithmetic progression and a,b+8,c+64 are in geometric progression. If the arithmetic mean of a, b, c is k, then 3/13k is equal to

Answer 1

To solve this problem, we can set up and solve equations using the given information. Once we find the values of a, b, c, r, and d, we can calculate the arithmetic mean and find 3/13k.

Step-by-step explanation:

To solve this problem, we can use the given information to create equations and solve them. Let's denote the common ratio of the geometric progression as r and the common difference of the arithmetic progression as d. Using these variables, we can write the equations:

1. a = r
2. b+8 = a+d
3. c = a+d+d = a+2d
4. b+8+64 = (b+8)r = br+8r+64
5. c+64 = (c+64)r = (a+2d)+64r = ar+8r+128r+64

Let's simplify these equations and solve them:

1. r = a
2. 2a+d = a+8
3. a+2d = a+2d
4. br+8r+64 = (b+8)r
5. ar+8r+128r+64 = (a+2d)+64r

By solving these equations simultaneously, we can find the values of a, b, c, r, and d. Once we have those values, we can calculate the arithmetic mean of a, b, c and find 3/13k.