Question

If a,b,c are in A.P then show that I/bc,1/ca,1/ab are also in Ap​

Answer 1

Answer:

a, b, c are in ap

Step-by-step explanation:

So

b - a = d (( x))

c - b = d (( y))

in second series

$\frac{1}{ca} - \frac{1}{bc} = \frac{b - a}{abc} \\ so. \: using \:x \: \: b - a = d \: \: we \: \: get \\ \frac{1}{ca} - \frac{1}{bc} = \frac{d}{abc} \: \: \: \: eq1$

$\frac{1}{ab} - \frac{1}{ca} = \frac{c - b}{abc} \\ \\ \\ so. \: using \: y \: c - b \: \: = d \: \\ \frac{1}{ab} - \frac{1}{ca} = \frac{c - b}{abc} = \frac{d}{abc} \: \: \: eq2$

Now eq 1 = eq 2

actually eq 1 and 2 are their common difference.

so hence common difference is same so no. are in ap.