If a, b, c are in AP; b, c, d are in GP; c, d, e are in HP, Prove that a, c, e are in GP.

Let us note down the three given conditions

a, b, c are in AP => 2b = a + c                                                       (1)
b, c, d are in GP => c² = bd                                                           (2)
c, d, e are in HP                                                                            
or
1/c, 1/d, 1/e are in HP => 2/d = 1/c + 1/e                                       (3)

We have to prove that a, c, e are in GP or, in mathematical terms, c² = ae.

We'd first solve by the simple, straight forward approach which is commonly the easiest to think of. (2) is already c² = bd. b can be expressed in terms of a and c from (1) and d can be expressed in terms of c and e from (3). In this way, the expression (2) will consist of the variables c, a, e - just the kind of expression we want to prove.

From (1), b = (a + c)/2

From (3), 2/d = (c + e)/ce => d = 2ce/(c + e)

Substituting these in (2), we get

c² = (a + c)/2 * 2ce/(c + e)
=> c² = ce(a + c)/(c + e)
=> c² (c + e) = ce (a + c)
=> c³ + ec² = ace + ec²
=> c³ = ace
=> c³ - ace = 0
=> c(c² - ae) = 0
=> c = 0, c² = ae

For c = 0, all a, b, c, d, and e become 0, and c² = ae is automatically proved. The other case also makes c² = ae. Hence, c is always equal to ae.

THE NOT SO OBVIOUS APPROACH - This might also strike some minds. We can see that (1) is 2b = a + c. Now, the expression could be divided by bd in the LHS and c² in the RHS (because both of them are equal) to make its LHS equal to the LHS of (3) and hence the RHS of (3). The RHS of (1) will then consist of terms a and c only, just like the RHS of (3). We have done it here - 

2b = a + c
=> 2b/bd = a/bd + c/bd              (dividing both sides by bd)
=> 2/d = a/c² + c/c²                    (because (2), c² = bd)
=> 2/d = a/c² + 1/c
=> 1/c + 1/e = a/c² + 1/c            (because (3), 2/d = 1/c + 1/e)
=>  1/e = a/c²
=> c² = ac

Hence Proved.