If a,b,c are in A.P. and a,b,d are in G.P. ,prove that a,a-b,d-c are in G.P.

 a,b ,c are in ap 

so 2b =a +c

 and a,b ,d are in GP

  b2 = ad

  now  a, a-b , d-c

  a-b / a = d-c / a-b

  (a-b)2 =  a(d-c)

  a2 +b2 -2ab = ad - ac 

  a2 + ad -  a(a+c)  = ad - ac 

  a2 - a2 - ac = -ac

  ac =ac TRUE ......

hence a, a-d , d-c are in GP

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As  a,b,c are in A.P. 

therefore , 2b = a + c 

And a , b and d are in G.P. 

therefore b^2 = ad 

b^2  - ac = ad - ac

a ^ 2 + b ^ 2 -a^2 - ac = ad - ac

a ^ 2  + b ^ 2 - a (a+c) = ad - ac

a ^ 2 + b ^ 2 - 2ab = a (d -c)

(a - b) ^ 2 = a (d-c)

Thus , a ,a-b , d-c are in G.P .

Thanks Bye!!!!!!

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