If 2x, x + 8, 3x + 1 are in A.P.,then the val | vedclass

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(C) Given that the terms $2x, x + 8, 3x + 1$ are in $A.P.$For any three terms $a, b, c$ to be in $A.P.$,the middle term must be the arithmetic mean of

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$5$

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(C) Given that the terms $2x, x + 8, 3x + 1$ are in $A.P.$For any three terms $a, b, c$ to be in $A.P.$,the middle term must be the arithmetic mean of the other two,i.e.,$b = \frac{a + c}{2}$.Applying this to the given terms:$x + 8 = \frac{2x + (3x + 1)}{2}$Multiply both sides by $2$:$2(x + 8) = 5x + 1$$2x + 16 = 5x + 1$Rearranging the terms to solve for $x$:$16 - 1 = 5x - 2x$$15 = 3x$$x = \frac{15}{3}$$x = 5$

If $2x, x + 8, 3x + 1$ are in $A.P.$,then the value of $x$ will be

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