Question

If a ,b ,c are in AP then show that (b+c)^2-a^2 , (c+a)^2-b^2 , (a+b)^2-c^2 are also in AP
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Answer 1

Answer:

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Step-by-step explanation:

Given: a,b,c are in AP

Since,a,b,c are in AP, we have a+c=2b....(i)

Now,(b+c)2−a2,(c+a)2−b2,(a+b)2−c2 will be in A.P.

If (b+c−a)(b+c+a),(c+a−b)(c+a+b),(a+b−c)(a+

b+c) are in AP

i.e. if b+c−a,c+a−b,a+b−c are in AP

[dividing by(a+b+c)]

if (b+c−a)+(a+b−c)=2(c+a−b)

If 2b=2(c+a−b)

if b=c+a−b

if a+c=2b which is true by (i)

Hence,(b+c)2−a2,(c+a)2−b2,(a+b)2−c2  are in A.P

Answer 2

Answer:

Given: a,b,c are in AP

Since,a,b,c are in AP, we have a+c=2b....(i)

Now,(b+c)

2

−a

2

,(c+a)

2

−b

2

,(a+b)

2

−c

2

will be in A.P.

If (b+c−a)(b+c+a),(c+a−b)(c+a+b),(a+b−c)(a+

b+c) are in AP

i.e. if b+c−a,c+a−b,a+b−c are in AP

[dividing by(a+b+c)]

if (b+c−a)+(a+b−c)=2(c+a−b)

If 2b=2(c+a−b)

if b=c+a−b

if a+c=2b which is true by (i)

Hence,(b+c)

2

−a

2

,(c+a)

2

−b

2

,(a+b)

2

−c

2

are in A.P

Step-by-step explanation:

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