Let {an} be an increasing arithmetic sequence, the sum of the first three terms is 12, and the product of the first three terms is 48, then its first term is () A. 1B. 2C. 4D. 6

Let {an} be an increasing arithmetic sequence, the sum of the first three terms is 12, and the product of the first three terms is 48, then its first term is () A. 1B. 2C. 4D. 6

Let the first three terms of {an} be A1, A2, A3, then from the properties of the arithmetic sequence, we can get a1 + a3 = 2A2, ∵ a1 + A2 + a3 = 3a2 = 12, the solution is A2 = 4, from the meaning of the question, we can get a1 + A3 = 8a1a3 = 12, the solution is A1 = 2A3 = 6 or A1 = 6a3 = 2, ∵ {an} is an increasing arithmetic sequence, ∵ A1 = 2, A3 = 6, so choose B

Insert three numbers between 12 and 60, make them form an arithmetic sequence with the two numbers, and find the three numbers

Let these three numbers be a, B, C, then 12, a, B, C, 60 constitute the arithmetic sequence. B is the median of 12 and 60, so 2B = 12 + 60, so B = 36. Similarly, a is the median of 12 and 36, so a = 24, similarly, C = 48

The sum of three numbers is 21 and the product is 91. How many are these three numbers?
It's best to use the formula when you are in a hurry

The sum of the three numbers is 21, and the number in the middle is 21 △ 3 = 7
The product is 91. Let these three numbers be 7-x, 7, 7 + X
(7-x )(7+x) = 91 ÷7
49 - x² = 13
x² = 36
x = ± 6
These three numbers are 1, 7, 13
Or 13, 7, 1