Browse Questions

# Among 5 numbers a, b, c, d and e the relationships are: a, b, c are in AP and b, c, d are in GP and c, d, e are in HP. Find b, c, d if a = 2 and e = 18.

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

Toolbox:
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

Toolbox:
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