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(D) Given that $a, 1, b$ are in arithmetic progression $(AP)$.Therefore,$1 = \frac{a+b}{2}$,which implies $a+b = 2$ $......(1)$.Given that $1, a, b$ a

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$-2, 4$

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(D) Given that $a, 1, b$ are in arithmetic progression $(AP)$.Therefore,$1 = \frac{a+b}{2}$,which implies $a+b = 2$ $......(1)$.Given that $1, a, b$ are in geometric progression $(GP)$.Therefore,$a^2 = 1 	imes b$,which implies $b = a^2$ $......(2)$.Substituting the value of $b$ from equation $(2)$ into equation $(1)$:$a + a^2 = 2$$a^2 + a - 2 = 0$Factoring the quadratic equation:$(a+2)(a-1) = 0$So,$a = -2$ or $a = 1$.If $a = 1$,then $b = (1)^2 = 1$. However,the problem states $a 
eq b$,so we reject this case.If $a = -2$,then $b = (-2)^2 = 4$.Thus,the values are $a = -2$ and $b = 4$.

Terms $a, 1, b$ are in arithmetic progression and terms $1, a, b$ are in geometric progression. Find $a$ and $b$ (given $a \neq b$).

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