Question

guys,solve this plzzz.......

Answer 1

if a,b,c are in AP, 
b-a = c-b
to prove 1/bc, 1/ca, 1/ab are in AP, we need to show 1/ca - 1/bc = 1/ab - 1/ca
LHS:
1/ca - 1/bc = (b-a)/abc   
RHS:
1/ab - 1/ca = (c-b)/abc

but we know that (b-a) = (c-b)
thus LHS = RHS.
Hence 1/bc, 1/ca, 1/ab are in AP.

Note: Calculation of 1/ca - 1/bc
You need to first take the LCM of ca and bc which is abc and do the calculation. I have calculated like this above.

Or 
\frac{1}{ca} - \frac{1}{bc} = \frac{bc-ca}{abc^{2}} = \frac{c(b-a))}{abc^{2}} = \frac{b-a}{abc}ca1​−bc1​=abc2bc−ca​=abc2c(b−a))​=abcb−a​  
Here instead of taking LCM, i have multiplied the terms, which is fine. You will get the same answer.

Answer 2

If a, b, c are in AP.

Then,

b-a = c-b = common difference
2b = c+a
b = (c+a)/2

Let's substitute this value of b in the next AP series,
which are,
$\frac{1}{bc} , \frac{1}{ac} , \frac{1}{ab}$


which after substitution will become,

$\frac{2}{(c + a)c} , \frac{1}{ca}, \frac{2}{a(a + c)}$



Now if the 2nd term - 1st term = 3rd term -2nd term.
Then this series is in AP.


Let's check that,

$\frac{1}{ca} - \frac{2}{c(a + c)} \\ = \frac{c(a + c) - 2ac}{a {c}^{2}(a + c) } \\ = \frac{ {c}^{2} - ac }{a {c}^{2} (a + c)} \\ \frac{c(c - a)}{a {c}^{2} (a + c)} = \frac{ - 1}{ac}$

Similarly,
$\frac{2}{a(a + c)} - \frac{1}{ac} \\ = \frac{ - 1}{ac}$

The common difference here is constant.

Hence, It's in AP.