Let 3, a, b, c be in A.P. and 3, a-1, b+1, c+ | vedclass

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(D) Given $3, a, b, c$ are in $A.P.$,let the common difference be $d$. Then $a = 3+d, b = 3+2d, c = 3+3d$.Given $3, a-1, b+1, c+9$ are in $G.P.$,we su

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$11$

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(D) Given $3, a, b, c$ are in $A.P.$,let the common difference be $d$. Then $a = 3+d, b = 3+2d, c = 3+3d$.Given $3, a-1, b+1, c+9$ are in $G.P.$,we substitute the values:$3, (3+d)-1, (3+2d)+1, (3+3d)+9 \Rightarrow 3, 2+d, 4+2d, 12+3d$ are in $G.P.$For a $G.P.$,the ratio of consecutive terms is constant,so $(2+d)^2 = 3(4+2d)$.$4 + 4d + d^2 = 12 + 6d \Rightarrow d^2 - 2d - 8 = 0$.$(d-4)(d+2) = 0$,so $d = 4$ or $d = -2$.Case $1$: If $d = 4$,then $a = 7, b = 11, c = 15$. The sequence is $3, 6, 12, 24$,which is a $G.P.$The arithmetic mean of $a, b, c$ is $\frac{7+11+15}{3} = \frac{33}{3} = 11$.Case $2$: If $d = -2$,then $a = 1, b = -1, c = -3$. The sequence is $3, 1, -1, 3$,which is not a $G.P.$ (as $1/3 
eq -1/1$).Thus,the only valid solution is $11$.

Let $3, a, b, c$ be in $A.P.$ and $3, a-1, b+1, c+9$ be in $G.P.$ Then,the arithmetic mean of $a, b,$ and $c$ is:

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