1)a,b,c are three unequal numbers in GP Then a:b:c=2)If a,b,c,d are in HP, then ab+bc+cd = ?

ans 1.:-

1:r:r^2

ans2:-

i dont know.do by ur own.

Let the terms of the H.P are (1/a)= a, (1/(a+d))= b , (1/(a+2d)) = c and (1/(a+3d)) =d where a, d are the first term and the common differences  respectively. 

Then ab+bc+bd =

= 1/a(a+d)+1/(a+d)(a+2d)+1/(a+2d)(a+3d)

= ((a+2d)(a+3d)+(a)(a+3d)+(a)(a+d))/(a(a+d)(a+2d)(a+3d))

= ((a^2+5ad+6d^2)+(a^2+3ad)+(a^2+ad))/(a(a+d)(a+2d)(a+3d))

= (3a^2+9ad+6d^2)/(a(a+d)(a+2d)(a+3d))

= ((3)(a+d)(a+2d))/(a(a+d)(a+2d)(a+3d))

= 3/a(a+d)

Since 1/(a+d) = d and 1/a =a 

Then =  3ad

A Slight correction in last few lines

=3/(a)(a+3d)

since (a+3d)= d

=3ad

1)ar where a is first term and ris common ratio        

2)3ad where a is first term and d is common difference