2024/023) If a, b, c be in Arithmetic Progression, then the value of $(a+2b-c)(c+2b-a)(a+2b+c)$ is nabc. Find n

 We have a,b,c are in AP so 

$2b = a+ c$

Hence $a+2b -c = a + a+c -c = 2a\cdots(1)$

$c + 2b -a = c + a + c - a = 2c\cdots(2)$

$a + 2b + c= 2b + a +c = 2b+2b = 4b\cdots(3)$ 

Multiplying (1) (2) and (3) we get $(a+2b-c)(c+2b-a)(a+2b+c)= 16abc$

Hence n = 16