It is known that three positive numbers a, B and C satisfy a 2, B 2 and C 2 as arithmetic sequence. It is proved that 1A + B, 1A + C and 1b + C are also arithmetic sequence

It is known that three positive numbers a, B and C satisfy a 2, B 2 and C 2 as arithmetic sequence. It is proved that 1A + B, 1A + C and 1b + C are also arithmetic sequence

∵ if three positive numbers a, B, C satisfy A2, B2, C2 as arithmetic sequence, ∵ A2 + C2 = 2B2, ∵ 1A + B + 1b + C = a + 2B + C (B + C) (a + b), ∵ if a + 2B + C (B + C) (a + b) = 2A + C holds, then it is equivalent to 2Ab + 2B2 + 2BC + 2Ac = A2 + 2Ab + AC + AC + 2BC + C2, and it is simplified to 2B2 = A2 + C2, this formula holds, ∵ conclusion holds. It is proved that 1A + B, 1A + C, 1b + C also becomes arithmetic sequence

A mathematical problem about arithmetic sequence
In the arithmetic sequence with 2N-1 terms, if the sum of all odd terms is 165 and the sum of all even terms is 150, what is the value of N?

2N-1 is obviously an odd number. There are n odd terms, n-1 even terms. The difference between two adjacent odd terms is 2D. The sum of n terms of odd terms is Na1 + n (n-1) 2D / 2 = n [A1 + (n-1) D]. ① the sum of n-1 even terms is (n-1) A2 + (n-1) (n-2) 2D / 2. A2 = a1 + D (n-1) A2 + (n-1) (n-2)

If the sum of three integers is 30 and the square sum of three integers is 318, find the three integers

Let X in the middle be X-D, x, x + D, respectively
x+d+x+x-d=3x=30
x=10
x^2+(x+d)^2+(x-d)^2=3x^2+2d^2=300+2d^2=318
d^2=9
d=3
So it's 7, 10, 13