Question

if a,b,c are in ap ,show that

Answer 1

\[a(\frac{ 1 }{ b }+\frac{ 1 }{ c }),b(\frac{ 1 }{ c }+\frac{ 1 }{ a }),c(\frac{ 1 }{ a }+\frac{ 1 }{ b })\] also in AP

Answer 2

wwe123 what does ap mean? I am guessing arithmetic progression?

Answer 3

yes @phi

Answer 4

ap is arithmetic progression?

Answer 5

take the differences between terms and show you get the same difference

Answer 6

differences between terms ??

Answer 7

yes, you have 3 terms
\[ a(\frac{ 1 }{ b }+\frac{ 1 }{ c }),b(\frac{ 1 }{ c }+\frac{ 1 }{ a }),c(\frac{ 1 }{ a }+\frac{ 1 }{ b }) \]

Answer 8

to show they are in an arithmetic progression, show that the difference between the 1st and 2nd equals the difference between the 2nd and 3rd

Answer 9

how ?? @phi

Answer 10

using tedious algebra?

Answer 11

the first term can be written as
\[ \frac{a(c+b)}{bc} = \frac{a^2c+a^2b}{abc}\]
the 2nd term can be written as
\[ \frac{ab^2+b^2c}{abc} \]
subtract them:
\[ \frac{ab^2+b^2c-a^2c-a^2b}{abc}\]
now simplify, and use the fact that a,b,c are in AP so b-a= d and c-b=d
(d is the common difference between the terms)

do the same for the 2nd and 3rd terms. You will get the same difference

Answer 12

we have to take a,b,c and multiply or divided by some to get (me first comment)

Answer 13

the numerator is
\[ (ab^2-a^2b) + (b^2c - a^2c) = ab(b-a) + c(b^2-a^2) \]
\[ = ab(b-a) + c(b-a)(b+a)= (b-a)(ab+bc+ac) = d(ab+bc+ac) \]

Answer 14

what are u trying to do @phi

Answer 15

An example of an arithmetic progression is
1,4,7
if we subtract consecutive terms: (4-1)= 3 and (7-4)= 3
we get a common difference of 3

the problem tells us
a,b,c are in AP, so (b-a)= d and c-b= d

we want to show that the difference between consecutive terms in the new series gives a common difference.

I found the difference between the 2nd and 1st term to be
\[ \frac{d(ab+ac+bc}{abc} \]

now what is left is to show the difference between the 2nd and 3rd terms is that same expression

Answer 16

@phi in this question we have not to find but we have to prove

Answer 17

one way to show a sequence of terms are an AP is to show they have a common difference

Answer 18

if i may suggest an alternative method,
we need to prove , ....., ...., .... are in AP
so show, a(1/b+1/c) + c (1/a+1/b) = 2 b[1/c+1/a]
little less algebraic cumbersome.....

Answer 19

but how to show @hartnn

Answer 20

not too difficult, u just need to use , a+c=2b

Answer 21

plz show :(

Answer 22

atleast try once...

Answer 23

or follow phi...

Answer 24

thank all