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Q.

Consider three sequences an A.P., a G.P. and a H.P. as follows 
 a,A1,A2,A3.......A9,b(A.P.)
 a,G1,G2,G3,....G9,b(G.P.);
 a,H1,H2,H3,......H9,b(H.P.)
If  a,b are both roots of the equation  x2(p+1)x+16=0,pR , then the value of  r=14ArH10rG102r  is 

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a

16

b

64

c

49

d

412

answer is D.

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Detailed Solution

 A1H9G8 A2H8G6 A3H7G4 A4H6G2 
Now  A1H9=A2H8=A3H7=ab=16
       Expression  =164×G2G4G6G8
Now,  G2G8=G4G6=ab=16
                   164×162=412

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