Answers

2014-07-22T21:36:06+05:30
Let a/x, a, ax be three terms of GP, whose common ratio is x.
a/q, 
         (a/x)^2+a+(ax)^2 = 91  ------------eqn(1)
         a/x +a+ax = 13  --------eqn(2)
   or, a(x^2+x+1) = 13x
we know that ;
            (a/x+a+ax)^2 = (a/x)^2+a^2+(ax)^2+2a^2/x+2a^2x+2a^2
     now put the values of (a/x+a+ax)  &  (a/x)^2+a^2+(ax)^2 from eqn (1) & (2).
        or, 13^2 = 91+2a^2(1/x+x+1)
        or,  2a^2(x^2+x+1)/x = 169-91
        or, 2a^2(x^2+x+1) = 68x
put the value of x  from (1)
       or, 2a^2(x^2+x+1) = 68a(x^2+x+1)/13
       or, 2a=68/13
      or, a=34/13
now we can find the value of x,when put the value of a in (2)
so series of GP 34/13x, 34/13, 34x/13
1 1 1
169-91=78....
ok but method is right