Let a, b in R be such that a, a + 2b, 2a + b | vedclass

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(C) Given that $a, a + 2b, 2a + b$ are in $A.P.$Therefore,$2(a + 2b) = a + (2a + b)$$2a + 4b = 3a + b$$a = 3b$  --- $(1)$Also,$(b + 1)^2, ab + 5, (a +

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$4$

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(C) Given that $a, a + 2b, 2a + b$ are in $A.P.$Therefore,$2(a + 2b) = a + (2a + b)$$2a + 4b = 3a + b$$a = 3b$  --- $(1)$Also,$(b + 1)^2, ab + 5, (a + 1)^2$ are in $G.P.$Therefore,$(ab + 5)^2 = (b + 1)^2(a + 1)^2$ --- $(2)$Substituting $(1)$ into $(2)$:$(3b^2 + 5)^2 = (b + 1)^2(3b + 1)^2$Taking the square root on both sides:$3b^2 + 5 = \pm(b + 1)(3b + 1)$Case $1$: $3b^2 + 5 = (b + 1)(3b + 1)$$3b^2 + 5 = 3b^2 + 4b + 1$$4b = 4 \Rightarrow b = 1$Since $a = 3b$,$a = 3(1) = 3$.Thus,$a + b = 3 + 1 = 4$.Case $2$: $3b^2 + 5 = -(b + 1)(3b + 1)$$3b^2 + 5 = -(3b^2 + 4b + 1)$$6b^2 + 4b + 6 = 0$$3b^2 + 2b + 3 = 0$The discriminant $D = 2^2 - 4(3)(3) = 4 - 36 = -32 Since $D Therefore,the only real solution is $a = 3, b = 1$,and $a + b = 4$.

Let $a, b \in R$ be such that $a, a + 2b, 2a + b$ are in $A.P.$ and $(b + 1)^2, ab + 5, (a + 1)^2$ are in $G.P.$ then $(a + b)$ equals

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