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

2 Answers
Ashley H
Aug 12, 2018

For, #a,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

Ratnaker Mehta
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!

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