$(a)\;4,6,9\qquad(b)\;2,6,18\qquad(c)\;4,6,8\qquad(d)\;-2,-6,18$

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- Answer : (a) 4,6,9

since a,b,c in AP

$b=\frac{a+c}{2}=\frac{2+c}{2}\qquad(a=2)$

c,d,e are in HP

$d=\frac{2ce}{c+e}=\frac{36c}{c+18}\qquad(e=18)$

b,c,d are in GP

$c^2=bd=(\frac{2+c}{2})(\frac{36c}{c+18})$

$2c^2+36c=72+36c$

$2c^2=72,\;c^2=36,\;c=6$

$with \;c=6,\;b=\frac{2+c}{2}=\frac{2+6}{c}=4$

$d=\frac{36c}{c+18}=\frac{36*6}{6+18}=9$

b,c,d=4,6,9

$with\;c=-6,b=\frac{2+c}{2}=\frac{2-6}{2}=-2$

$d=\frac{36*(-6)}{(-6)+18}=-18$

b,c,d=-2,-6,-18

since the 2nd option is not in the choices.answer is (a) b,c,d=4,6,9

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- Answer : (a) 4,6,9

since a,b,c in AP

$b=\frac{a+c}{2}=\frac{2+c}{2}\qquad(a=2)$

c,d,e are in HP

$d=\frac{2ce}{c+e}=\frac{36c}{c+18}\qquad(e=18)$

b,c,d are in GP

$c^2=bd=(\frac{2+c}{2})(\frac{36c}{c+18})$

$2c^2+36c=72+36c$

$2c^2=72,\;c^2=36,\;c=\pm6$

$with \;c=6,\;b=\frac{2+c}{2}=\frac{2+6}{c}=4$

$d=\frac{36c}{c+18}=\frac{36*6}{6+18}=9$

b,c,d=4,6,9

$with\;c=-6,b=\frac{2+c}{2}=\frac{2-6}{2}=-2$

$d=\frac{36*(-6)}{(-6)+18}=-18$

b,c,d=-2,-6,-18

since the 2nd option is not in the choices.answer is (a) b,c,d=4,6,9

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