Question

if a,b,c are in A.P then (a+2b-c)(2b+c-a)(c+a-b)​

Answer 1

Answer:

4abc

Step-by-step explanation:

b-c=c-b

= 2b=a+c

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

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

= (2a) (2c) (b)

= 4abc

Answer 2

$\huge{ \bigstar}\huge{\mathcal{ \overbrace{ \underbrace{ \red{ \fbox{ \green{ \blue{A} \pink{n} \red{s} \green{w} \purple{e} \blue{r}}}}}}}} \huge{ \bigstar}$

$\red{ \underline{ \underline{ \tt{Question}}}} : -$

If a, b & c are in A.P. Then prove that (a + 2b - c)(2b + c - b)(c + a - b) = 4abc

$\tt{ \underline{ \underline{ \purple{Solution}}}} : -$

Since, a, b & c are in A.P.

So, a + c = 2b ———(i)

L.H.S.

$\: \: \: \:(a+ 2b- c)(2b+c-a)(c + a - b) \\ = (a + a + \cancel c - \cancel{c})( \cancel{a} + c + c - \cancel a)(2b - b) \\ \\ \:\:\:\: [\bf{ \orange{From \: equation - (i)}}] \\ \\ = 2a \times 2c \times b \\ = 4abc$