Discussion
Let \(a_{1}, a_{2}, 6, a_{4}\) be in G.P. Then the value of \((6a_{1} + 6a_{2} + a_{1}a_{4} + a_{2}a_{4})^{2}\) is equal to
Answer: A
The value is \((6 + a_{2})^{4}\).
Since \(a_{3} = 6\), the bracket factorises:
\[ \begin{aligned} 6a_{1} + 6a_{2} + a_{1}a_{4} + a_{2}a_{4} &= (a_{1} + a_{2})(6 + a_{4}) \\ &= (a_{1} + a_{2})(a_{3} + a_{4}) \end{aligned} \]
In a G.P., \(a_{2} = a_{1}r\), \(a_{3} = a_{2}r\) and \(a_{4} = a_{3}r\), so \(a_{3} + a_{4} = r^{2}(a_{1} + a_{2})\) and \(a_{2} + a_{3} = r(a_{1} + a_{2})\). Hence
\[ \begin{aligned} (a_{1} + a_{2})(a_{3} + a_{4}) &= r^{2}(a_{1} + a_{2})^{2} \\ &= (a_{2} + a_{3})^{2} \end{aligned} \]
Squaring gives \((a_{2} + a_{3})^{4} = (6 + a_{2})^{4}\). The other options pair 6 with the wrong neighbour.
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