Final answer:
To prove that a^(n+1) = ab when a1 is the first geometric mean between a and b, we use properties of geometric progressions, find the common ratio, and manipulate the terms accordingly to arrive at the desired relationship.
Step-by-step explanation:
The question seems to have a typo but interprets as showing that if a1 is the first term in a series of n geometric means between a and b, then the relationship a^(n+1) = ab holds. To proceed with the solution, we recall that in a geometric progression, any term can be found using the formula Tn = a * r^(n-1), where Tn is the nth term, a is the first term, and r is the common ratio. If a1 is the first geometric mean, then a1 = a * r, and the last term (which is the nth geometric mean) would be an = a * r^n. Since an is also equal to b, we can say that b = a * r^n. Dividing the equation of b by that of a1, we get r^n = b/a. Taking both sides to the nth root, we find that r = (b/a)^(1/n). To find a^(n+1), we raise the first term a to the power of n+1, obtaining a^(n+1) = a * a^n. Substituting for r in the term an we found earlier, we get a^(n+1) = a * (b/a) = b, thus a^(n+1) = ab, which was to be proven. This validates the original statement in the question, showing that the product of the first term raised to the n+1 power and the last term in the sequence equals the product of the first and last terms.