Prove that the sum of n arithmetic means between a and b is n times their arithmetic mean.

You're on a roll — The solution provided in the image correctly demonstrates that the sum of n Arithmetic Means (A.Ms) between a and b is equal to n times their A.M. Explanation of the steps: 1. Identify the A.P. and its terms: Let A_1, A_2, , A_n be the n A.Ms between a and b. This forms an Arithmetic Progression (A.P.): a, A_1, A_2, , A_n, b. The first term of this A.P. is a_1 = a. The total number of terms in this A.P. is N = n+2. The last term is a_n+2 = b. 2. Find the common difference (d): Using the formula for the k-th term of an A.P., a_k = a_1 + (k-1)d: For the (n+2)-th term: a_n+2 = a_1 + ((n+2)-1)d b = a + (n+1)d Solving for d: d = (b-a)/(n+1) This matches the derivation in the image. 3. Calculate the sum of the n A.Ms (A_1 + A_2 + + A_n): The sum of an A.P. with k terms is S_k = (k)/(2)(first term + last term). Here, we are summing the n terms A_1, , A_n. The first term of this sub-sequence is A_1, and the last term is A_n. So, the sum is S_n = (n)/(2)(A_1 + A_n). Now, express A_1 and A_n in terms of a and d: A_1 is the second term of the main A.P., so A_1 = a_1 + d = a + d. A_n is the (n+1)-th term of the main A.P., so A_n = a_1 + nd = a + nd. Substitute these into the sum formula: S_n = (n)/(2)((a+d) + (a+nd)) S_n = (n)/(2)(2a + (n+1)d) This matches the image. 4. Substitute the value of d into the sum: Substitute d = (b-a)/(n+1) into the expression for S_n: S_n = (n)/(2)(2a + (n+1)((b-a)/(n+1))) The (n+1) terms cancel out: S_n = (n)/(2)(2a + (b-a)) S_n = (n)/(2)(a+b) This matches the image. 5. Relate to the Arithmetic Mean: The Arithmetic Mean (A.M.) between a and b is (a+b)/(2). From the previous step, we have: S_n = n ((a+b)/(2)) S_n = n × (A.M. between a and b) This completes the proof. The solution provided in the image is correct. What's next?