Question

If $p, q, r$ are in A.P., $a$ is G.M. between $p$ and $q$, and $b$ is G.M. between $q$ and $r$, then $a^2, q^2, b^2$ are in:

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Hint:

For three terms to be in A.P., the sum of the first and third must equal twice the middle term.

Answer 1

The Correct Option is A

Step 1: Given that $p, q, r$ are in A.P., we can write: \[ q - p = r - q = d \] where $d$ is the common difference.
Step 2: Since $a$ is the geometric mean (G.M.) between $p$ and $q$: \[ a = \sqrtpq \] Step 3: Since $b$ is the geometric mean between $q$ and $r$: \[ b = \sqrtqr \] Step 4: We need to check whether $a^2, q^2, b^2$ are in A.P.
From $a = \sqrtpq$, we get $a^2 = pq$.
From $b = \sqrtqr$, we get $b^2 = qr$.
We already have $q^2$ from the middle term.
Step 5: For numbers $x, y, z$ to be in A.P., we must have: \[ 2y = x + z \] Here, $x = a^2 = pq$, $y = q^2$, $z = b^2 = qr$.
Step 6: Check the condition: \[ a^2 + b^2 = pq + qr \] Factor $q$: \[ pq + qr = q(p + r) \] Since $p, q, r$ are in A.P., $p + r = 2q$.
Thus: \[ pq + qr = q(2q) = 2q^2 \] This satisfies the A.P. condition: \[ a^2 + b^2 = 2q^2 \] Therefore, $a^2, q^2, b^2$ are in $\mathbfA.P.$.