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If a,b,c,da,b,c,d are in Arithmetic progession (AP,) and a,c,da,c,d are in Geometric Progession (GP) then Prove that
a^2-d^2a2d2 = 3(b^2-ad)3(b2ad)

2 Answers
Aug 12, 2018

For, a,b,c,da,b,c,d in A.P we can write

2b=a+c....1 (as, c-b=b-a)

2c=b+d....2 (as, d-c=c-b)

Doing 1+2,

(b+c)=(a+d)....3

Doing 1-2

3(b-c)=(a-d)....4

And, for a,c,d in G.P we get,
c^2=ad... 5 (as, c/a=d/c)

L.H.S =

a^2-d^2

=(a+d)(a-d)

=(b+c)×3(b-c) (putting values from 3 & 4)

=3(b^2-c^2)

=3(b^2-ad) (from 5)

=R.H.S

Aug 12, 2018

Kindly refer to The Explanation.

Explanation:

Given that, a,b,c,d are in AP.

:. b" is the AM of "a and c".

:. b=(a+c)/2, or, a+c=2b..........(ast_1).

Similarly, b+d=2c.....................(ast_2).

Then, (ast_1)+(ast_2) rArr (a+d)+(b+c)=2(b+c),

or, (a+d)=(b+c)...............(star_1).

Also, (ast_1)-(ast_2) rArr (a-d)-(b-c)=2(b-c),

or, (a-d)=3(b-c)............(star_2).

Utilising (star_1) and (star_2), we have,

a^2-d^2=(a+d)(a-d)=(b+c){3(b-c)},

i.e., a^2-d^2=3(b^2-c^2).

Here, we use the fact that a,c,d are in GP, so that,

c/a=d/c, or, c^2=ad.

Therefore, a^2-d^2=3(b^2-c^2)=3(b^2-ad), as desired!