Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

If a, b, c are in arithmetic progression, then the value of (a + 2b - c)(2b + c - a)(a + 2b + c) is (1) 16abc (2) 4abc (3) 8abc (4) 3abc

Solution:

Since a,b,c are in AP, we know 2b = a+c

(a + 2b – c) = a+a+c-c = 2a (substitute 2b = a+c)

(2b + c – a) = a+c+c-a = 2c

(a + 2b + c) = (a+a+c+c = 2a+2c = 2(a+c) = 2(2b) = 4b

So (a + 2b – c)(2b + c – a)(a + 2b + c)= 2a(2c)(4b)

= 16abc

Hence option (1) is the answer.

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