If $a,b,c$ are in AP, $b,c,d$ are in GP and $c,d,e$ are in HP, then $a,c,e$ are in ($a\neq b\neq c\neq d\neq e>0\in\mathbb R$) If $a,b,c$ are in AP, $b,c,d$ are in GP and $c,d,e$ are in HP, then $a,c,e$ are in? (HP/GP/AP/None)($a\neq b\neq c\neq d\neq e&gt;0\in\mathbb R$)


So $a+c=2b\tag{1}$
$ce=d^2\tag2$
$\frac1c+\frac1e=\frac2d\tag3$
But I cannot see any pattern between the three numbers. Is there any?
Just write everything in terms of $a$ and $b$:
$$c = 2b -a$$
$$d = \frac{(2b-a)^2}{b}$$
$$\frac{1}{e} = \frac{2}{d} - \frac{1}{c}$$
Therefore, substituting $c$ and $d$ from above:
$$e = \frac{(2b-a)^2}{a}$$

Now it is easy to see that the sequence $a, c, e$ is a G.P:
$$(a,c,e) = \Big(a, 2b-a, \frac{(2b-a)^2}{a}\Big)$$
therefore
$$\sqrt{ae} = c.$$
Solution:
a,b,c are in AP.So,
b=(a+c)/2
or,2b=a+c
or,c=2b-a............(1)
also
b,c,d are in GP.so
c^2=bd
(2b-a)^2÷b=d.............(2)
and
c,d,e are in HP
d=2ce/(c+e)
or,c+e/ce=2/d
or,1/e+1/c=2/d
or,1/e=2/d-1/c..........(3)
from equation 1 2 and  3, we have
1/e=2b/(2b-a)^2 - 1/2b-a
or,e=(2b-a)^2/a......4
now if a,c,e are in GP
c^2=ae
substituting the value of c and e
(2b-a)^2=a.{(2b-a)^2}/a
therefore. (2b-a)^2=(2b-a)^2
so, a,c,e are in GP