Need some help in Progressions...?

Question

1. If 1/(b+a)+1/(b-c)=1/a+1/c, then a,b,c are in:

2. If a^2, b^2, c^2 are in harmonic progression, then (ab)^2,(ac)^2,(bc)^2 aree in:

Answer 1

1/(b+a) + 1/(b-c) = 1/a + 1/c,

then a, b and c are in H.P.

I don't know the proof of this, but I know the proof of its converse. if a, b, c are in H.P, then

1/(b+a) + 1/(b-c) = 1/a + 1/c.

proof: suppose a, b, c are in harmonic progression. then, 1/a, 1/b, 1/c are in ap. therefore,

2/b = 1/a + 1/c................(1).

since a, b and c form an hp,

a/c = (a-b) / (b-c)

cross multiplying,

1/c(a-b) = 1/a(b-c) = k..................(2)

                           = 2 / { ac - bc + ab - ac }

                          = 2 / b(a-c).

2 / b(a-c) = k

2/b = k(a-c)..............(3)

from, (2)

1/(b-c) = ck..................(4)

a/c = (a-b) / (b-c)

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

           = (a+b) / (b-c)

a/c = (a+b) / (b-c)

1/c(a+b) = 1/a(b-c) = k

1/(a+b) = ck...............(5)

(5) + (4) =>

1/(b+a) + 1/(b-c) = k (a-c)

                        = 2/b.   [ from (3) ]

therefore, from (1),

1/(b+a) + 1/(b-c) = 1/a + 1/c.

I think you can prove the original statement by solving for b.

2) a^2, b^2, c^2 are in H.P, then (ab)^2, (ac)^2, (bc)^2 are in A.P

proof:

a^2, b^2, c^2 ----- hp

=> 1/a^2, 1/b^2, 1/c^2 -------- ap

=> (bc)^2, (ca)^2, (ab)^2 ----- ap  [ multiplying throughout by (abc)^2 ].