Let x and y be real numbers such that x,x+2y,2x+y form an A.P. while the numbers (y+1)2,xy+5,(x+1)2 form a G.P. Find x+y ?
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Solution
The condition for A.P. 2(x+2y)=3x+y⇒x=3y (1) The condition for G.P. (xy+5)2=(x+1)2(y+1)2⇒(3y2+5)2=[(3y+1)(y+1)]2=0,by(1) ⇒(y−1)(3y2+2y+3)=0,∴y=1,−1+2√2i,−1−2√2i But given x,y are real ∴y=1 and x=3∴x+y=4