If 1ab - c-1bc - a- 1ca - b be in H-P- then a-b-c are also in H-P-

Taking reciprocals we are given that #160;a(b + c), b(c + a), c(a + b) are in A.P.Subtract ab + bc + ca from each bc, ca, ab are in A.P.Divide by abc,or ...

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Standard XII

Mathematics

Law of Reciprocal

If 1ab + ...

Question

If $\frac{1}{a (b + c)}, \frac{1}{b (c + a)}, \frac{1}{c (a + b)}$ be in H.P., then a,b,c are also in H.P.

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Solution

Taking reciprocals we are given that
a(b + c), b(c + a), c(a + b) are in A.P.
Subtract ab + bc + ca from each
-bc, -ca, -ab are in A.P.
Divide by - abc,
or $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. or a,b,c are in H.P.

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If 1a(b+c),1b(c+a),1c(a+b) be in H.P., then a,b,c are also in H.P.

Taking reciprocals we are given that a(b + c), b(c + a), c(a + b) are in A.P.Subtract ab + bc + ca from each-bc, -ca, -ab are in A.P.Divide by - abc,or 1a,1b,1c are in A.P. or a,b,c are in H.P.

If $\frac{1}{a (b + c)}, \frac{1}{b (c + a)}, \frac{1}{c (a + b)}$ be in H.P., then a,b,c are also in H.P.

Taking reciprocals we are given that
a(b + c), b(c + a), c(a + b) are in A.P.
Subtract ab + bc + ca from each
-bc, -ca, -ab are in A.P.
Divide by - abc,
or $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. or a,b,c are in H.P.