In a GP, first term is1. If 4T2+5T3 is minimum, then its common ratio is

Given, a=1 and 4T2+5T3  is minimum. Let r be the  common  ratio  of the  GP,  then∴ 4T2+5T3=4(ar)+5ar2 [∵a=1]⇒ 4r+5r2=f(r)      [say]....(i)⇒ r(4+5r)=f(r)     f(r)=0r(4+5r)=0⇒r=0,−4/5We know that, if a>0, quadratic expressionax2+bx+c has least value at  x=−b2aFrom  the  graph  it is clear that, minima  occurs  of point −25,−45.∴ r=−25.