Question

If a, b, c are in A.P., b, c, a are in G.P., then show that c, a, b are in H.P. ​

Answer 1

Answer:

We have been given that

a b c are in ap . Thus, we have

b=\frac{a+c}{2}...(i)b=

2

a+c

...(i)

b c d are in gp. Thus, we have

c^2=bd...(ii)c

2

=bd...(ii)

c d e are in hp. Thus, we have

d=\frac{2ec}{e+c}d=

e+c

2ec

Substituting the value of d from (ii)

\frac{c^2}{b}=\frac{2ec}{e+c}

b

c

2

=

e+c

2ec

Cancel, c from both sides of the numerator

\frac{c}{b}=\frac{2e}{e+c}

b

c

=

e+c

2e

Cross multiplying, we get

ec+c^2=2beec+c

2

=2be

Substituting the value of b from (i)

\begin{gathered}ec+c^2=2(\frac{a+b}{2})e\\\\ec+c^2=ae+ec\end{gathered}

ec+c

2

=2(

2

a+b

)e

ec+c

2

=ae+ec

Cancel ae both sides

c^2=aec

2

=ae

Thus, we can conclude that a, c, e are in GP

Answer 2

Step-by-step explanation:

SOLUTION

GIVEN

• a,b,c are in A.P
• So, 2b = a + c

• b,c,a are in hp
• So, c² = ab

a² = bc

a² + ac = bc + ac

a(a + c) = c(a + b)

• Put 2b = a + c in the equation

c = 2ab/(a+b)

Hence, c, a, b are in HP