Problem Solving of AP,GP,HP,AGP

AP ||GP

If x y z are in GP(x y z>1) then (1)/(2x+log x) , (1)/(4x+log y) , (1)/(6x+log z) are in 1) AP 2) GP 3) HP 4) AGP

Let the positive numebrs a,b,c,d be in A.P. Then abc,abd,acd,bcd re (A) not in A.P., G.P., H.P. (B) in A.P. (C) in G.P. (D) in H.P.

Intro Ap,GP,HP,Pro Summation of Series GP,AM-GM Inequality

If three positive numbers a,b and c are in AP, GP and HP as well, than find their values.

Let f(x) be a polynomial function of second degree.If f(a)=f(-a) and a,b,c are in A.P.P.then f'(a),f'(b),f'(c) are in 1) G.P.2 ) H.P.3 ) A.G.P.4 ) A.P.

AP || AGP

Let the positive numbers a,b,c,d be in AP. Then abc,abd,acd,bcd are (2001,1M) not in AP/GP/HP ( b ) in AP in GP (d) in HP

If a,b,c are in AP or GP or HP,then (a-b)/(b-c) is equal to

Let the positive numbers a ,b ,ca d nd be in the A.P. Then a b c ,a b d ,a c d ,a n db c d are a. not in A.P. /G.P./H.P. b. in A.P. c. in G.P. d. in H.P.