a,b,c are in the A.P then value of a-2b+c is
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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
Step-by-step explanation:
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