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If a²,b²,c² are in A.P, then prove that the following are in A.P.

1. 1/b+c,1/c+a,1/a+b

2. a/b+c,b/c+a,c/a+b

1
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1. 1/b+c,1/c+a,1/a+b

2. a/b+c,b/c+a,c/a+b

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Since are in A.P, the difference between consecutive terms is same :

Divide each side and get

In view of splitting into two partial fractions, rewrite numerators as

That means are in A.P.

For**part 2**, we may start with the result from part 1 :

are in A.P.

Multiplying each term by gives another A.P :

are in A.P.

are in A.P.

Subtracting from each term still gives another A.P :

are in A.P.

as desired.

Divide each side and get

In view of splitting into two partial fractions, rewrite numerators as

That means are in A.P.

For

are in A.P.

Multiplying each term by gives another A.P :

are in A.P.

are in A.P.

Subtracting from each term still gives another A.P :

are in A.P.

as desired.