Dear student Given: 5x-y, 2x+y and x+2y are in A P⇒22x+y=5x-y+x+2y⇒4x+2y=6x+y⇒y=2x 1and also, x-12,xy+1 and y+12 are in G P⇒xy+12= x-12y+12⇒x2y2+1+2xy=x2+1-2xy2+1+2y⇒x2y2+1+2xy=x2y2+x2+2x2y+y2+1+2y-2xy2-2x-4xy⇒x2+2x2y+y2+2y-2xy2-2x-6xy=0⇒x2+2x22x+2x2+22x-2x2x2-2x-6x2x=0⇒x2+4x3+4x2+4x-8x3-2x-12x2=0⇒-4x3-7x2+2x=0⇒-4x2-7x+2=0⇒4x2+7x-2=0On solving , we getx=14 and x=-2 not possible as per question So, x=14, y=2×14=12 x+y=14+12=34Note: There is some mistake in the question , it will be x>0 (going as per options)So correct option is aNote: When a,b,c are in A P2b=a+cand When a,b,c are in G Pb2=ac Regards

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Given : 5 x - y, 2 x + y and x + 2 y are in A . P \Rightarrow 2 (2 x + y) = 5 x - y + x + 2 y \Rightarrow 4 x + 2 y = 6 x + y \Rightarrow y = 2 x . . . (1) and also, {(x - 1)}^{2}, (xy + 1) and {(y + 1)}^{2} are in G . P \Rightarrow {(xy + 1)}^{2} = {(x - 1)}^{2} {(y + 1)}^{2} \Rightarrow x^{2} y^{2} + 1 + 2 xy = (x^{2} + 1 - 2 x) (y^{2} + 1 + 2 y) \Rightarrow x^{2} y^{2} + 1 + 2 xy = x^{2} y^{2} + x^{2} + 2 x^{2} y + y^{2} + 1 + 2 y - 2 {xy}^{2} - 2 x - 4 xy \Rightarrow x^{2} + 2 x^{2} y + y^{2} + 2 y - 2 {xy}^{2} - 2 x - 6 xy = 0 \Rightarrow x^{2} + 2 x^{2} (2 x) + {(2 x)}^{2} + 2 (2 x) - 2 x {(2 x)}^{2} - 2 x - 6 x (2 x) = 0 \Rightarrow x^{2} + 4 x^{3} + 4 x^{2} + 4 x - 8 x^{3} - 2 x - 12 x^{2} = 0 \Rightarrow - 4 x^{3} - 7 x^{2} + 2 x = 0 \Rightarrow - 4 x^{2} - 7 x + 2 = 0 \Rightarrow 4 x^{2} + 7 x - 2 = 0 On solving, we get x = \frac{1}{4} and x = - 2 (not possible as per question) So, x = \frac{1}{4}, y = 2 \times \frac{1}{4} = \frac{1}{2} x + y = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} Note : There is some mistake in the question, it will be x > 0 (going as per options) So correct option is (a) Note : When a, b, c are in A . P 2 b = a + c and When a, b, c are in G . P b^{2} = ac

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