Question

someone pls answer this question number 2.​

Answer 1

Answer:

TRUE

We have,

a

2

,b

2

,c

2

are in A.P

⇒b

2

−a

2

=c

2

−b

2

⇒(b−a)(b+a)=(c−b)(c+b)

⇒

(c+b)

(b−a)

=

(b+a)

(c−b)

⇒

(c+b)(c+a)

(b−a)

=

(b+a)(c+a)

(c−b)

⇒

(c+b)(c+a)

(b+c−c−a)

=

(b+a)(c+a)

(c+a−a−b)

⇒

c+a

1

−

b+c

1

=

a+b

1

−

c+a

1

b+c

1

,

c+a

1

,

a+b

1

are in A.P

b+c

a+b+c

,

c+a

a+b+c

,

a+b

a+b+c

are in A.P

b+c

a

+1,

c+a

b

+1,

a+b

c

+1 are in A.P

b+c

a

,

c+a

b

,

a+b

c

are in A.P

Hence, this is the answer.

Step-by-step explanation:

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