Given: a, b, c are in A.P as well as in G.P

To find: Whether (A) They are also in H.P or (B) Their reciprocals are in A.P or (C) Both (A) and (B) are true or (D) Both (A) and (B) are false.

Explanation:

Let us assume that:

a = b - d, b = a + d, and c = a * r

where d is the common difference and r is the common ratio.

As a, b, and c are in G.P, we have:

b^2 = ac

Substituting the values of a and c from above, we get:

(b-d)^2 = a * a * r

Simplifying the above equation, we get:

b^2 - 2bd + d^2 = a^2 * r

We also know that a and b are in A.P, which means:

a + d = b

Substituting this in the above equation, we get:

b^2 - 2b(b-a) + (b-a)^2 = a^2 * r

Simplifying the above equation, we get:

b^2 - 2ab + a^2 = a^2 * r

Canceling the common term 'a^2' from both sides, we get:

b^2 - 2ab + a^2 = b^2 * r

Simplifying the above equation, we get:

a^2 - 2ab(1-r) + b^2(1-r) = 0

As a and b are in A.P, we have:

(a+b)/2 = b - d/2

Substituting this in the above equation, we get:

(a-b)^2(1-r) = d^2

We know that a, b, and c are in G.P, which means:

c/b = b/a

Substituting the values of b and c from above, we get:

(a * r)/b = b/(a-d)

Cross-multiplying the above equation, we get:

b^2 = a * (a-d) * r

Substituting the value of b^2 from above in the equation (a-b)^2(1-r) = d^2, we get:

(a-b)^2(a-d) = d^2 * r

We can write the above equation as:

(a-b)^2 = d^2 * r/(a-d)

As a, b, and c are in A.P as well as in G.P, we have:

c/a = b/a - d/a

Substituting the values of b and c from above, we get:

(a * r)/a = (a+d)/a - d/a

Simplifying the above equation, we get:

r = (2d)/a

Substituting the value of r in the equation (a-b)^2 = d^2 * r/(a-d), we get:

(a-b)^2 = 4d^3/(a(a-d))

Taking the reciprocal of a, b, and c, we get:

1/a, 1/b, and 1/c are also in G.P.

Therefore, we have:

(1/b)^2 = 1/ac

Substituting the values of a