1-a- 1-b- and -1-c- are in A- P- show that - bc - ca and ab are also in A-P-

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Standard X

Mathematics

Arithmetic Mean

1/a, 1/b, an...

Question

$\frac{1}{a}, \frac{1}{b}, a n d \frac{1}{c},$ are in A. P. show that : bc , ca and ab are also in A.P.

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Solution

We have

1/b – 1/a = 1/c – 1/b

when we multiply by abc, then we get

ac – bc = ab – ac

ab + bc = 2ac

ac = (ab + bc)/2

Which is the condition of AP of the term, a, b and c.

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Similar questions

\frac{1}{a}, \frac{1}{b}

and

\frac{1}{c}

are in A.P. Show that: bc, ca and ab are also in A.P.

Q. If

a, b, c

are in A.P. then prove that

\frac{1}{b c}, \frac{1}{c a}, \frac{1}{a b}

are also in A.P

If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., prove that:
(i) $\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in A.P.
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Q. If

\frac{1}{a}, \frac{1}{b}, \frac{1}{c}

are in A.P., prove that
(i) bc,ca,ab are in A.P.
(ii) a(b+c), b(c+a), c(a+b) are in A.P

Q. If a, b, c are in A.P., then the following are also in A.P.

\frac{1}{b c}, \frac{1}{c a}, \frac{1}{a b}

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1a,1b, and 1c, are in A. P. show that : bc , ca and ab are also in A.P.

We have 1/b – 1/a = 1/c – 1/b when we multiply by abc, then we get ac – bc = ab – ac ab + bc = 2ac ac = (ab + bc)/2 Which is the condition of AP of the term, a, b and c.

$\frac{1}{a}, \frac{1}{b}, a n d \frac{1}{c},$ are in A. P. show that : bc , ca and ab are also in A.P.

We have

1/b – 1/a = 1/c – 1/b

when we multiply by abc, then we get

ac – bc = ab – ac

ab + bc = 2ac

ac = (ab + bc)/2

Which is the condition of AP of the term, a, b and c.