Question

If k is any positive real number and k^a,k^b,k^c are three consecutive terms of a gp then prove that a,b,c are in ap

Answer 1

If a b c are in AP then 2b=a+c by condition.

Answer 2

a, b, and c are in A. P.

Step-by-step explanation:

If k is a positive real number and $k^{a}, k^{b}, k^{c}$ are three consecutive terms of a G.P. then we can write

$[k^{b}] ^{2} = k^{a} \times k^{c}$ {Since we know that if x, y, and z are in G.P., then xz = y²}

⇒ $k^{2b} = k^{a + c}$ {Since, we know from the properties of exponents,  $[a^{b}] ^{c} = a^{bc}$ and $a^{b} \times a^{c} = a^{b + c}$}

⇒ 2b = a + c

Now, the condition for three numbers x, y, and z be in A.P. is x + z = 2y.

Therefore, a, b, and c are in A. P. (Proved)