Question

Three numbers a, b, and c are in G.P. If abc = 27 and a + c = 10, then a² + b² + c² =

• A. 81
• B. 82
• C. 91
• D. 92
• E. 99

Hint:

In GP, \(b^2 = ac\) and product of three terms = \(b^3\).

Answer 1

The Correct Option is C

Step 1: Concept:
• Let the three numbers in GP be: \[ a,\quad ar,\quad ar^2 \]
• Given: \[ abc = 27 \]
• So, \[ a \cdot ar \cdot ar^2 = a^3 r^3 = (ar)^3 \]
• Hence, \[ (ar)^3 = 27 \Rightarrow ar = 3 \]

Step 2: Calculation:
• From \(ar = 3\), we get: \[ a = \frac{3}{r} \]
• Given: \[ a + c = 10 \Rightarrow \frac{3}{r} + 3r = 10 \]
• Multiply both sides by \(r\): \[ 3 + 3r^2 = 10r \]
• Rearranging: \[ 3r^2 - 10r + 3 = 0 \]
• Factorizing: \[ (3r - 1)(r - 3) = 0 \]
• So, \[ r = 3 \quad \text{or} \quad r = \frac{1}{3} \]

Step 3: Final Answer:
• Case 1: \(r = 3\) \[ a = 1,\; b = 3,\; c = 9 \Rightarrow a^2 + b^2 + c^2 = 1 + 9 + 81 = 91 \]
• Case 2: \(r = \frac{1}{3}\) \[ a = 9,\; b = 3,\; c = 1 \Rightarrow a^2 + b^2 + c^2 = 81 + 9 + 1 = 91 \]
• Correct Option: (C)