Question

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

• A. $(6+a_{2})^{4}$
• B. $(6+a_{1})^{4}$
• C. $(6+a_{4})^{4}$
• D. $(6+a_{3})^{4}$
• E. $(6+2a_{2})^{4}$

Hint:

Math Tip: When a specific term in a G.P. is a known constant (like $a_3=6$), expressing all other terms relative to that constant using the common ratio '$r$' often simplifies complex algebraic expressions into basic rational functions.

Answer 1

The Correct Option is A

Concept:
Sequences and Series - Properties of Geometric Progressions.
Factorization by grouping terms.
Step 1: Factor the given expression.
The given expression inside the square is $6a_1 + 6a_2 + a_1a_4 + a_2a_4$.
Group the terms and factor them: $$ = 6(a_1 + a_2) + a_4(a_1 + a_2) $$ $$ = (6 + a_4)(a_1 + a_2) $$ So, the total expression is $[(6 + a_4)(a_1 + a_2)]^2$.
Step 2: Express terms using the common ratio $r$.
We are given the sequence: $a_1, a_2, 6, a_4$.
Notice that the third term $a_3 = 6$.
Let $r$ be the common ratio. We can write the other terms in relation to $a_3$:

• $a_4 = 6r$
• $a_2 = \frac{6}{r}$
• $a_1 = \frac{6}{r^2}$

Step 3: Substitute these values into the factored expression.
Substitute the terms into $(6 + a_4)(a_1 + a_2)$: $$ = (6 + 6r) \left( \frac{6}{r^2} + \frac{6}{r} \right) $$
Step 4: Simplify the expression.
Factor out the common terms: $$ = 6(1 + r) \cdot \left( \frac{6 + 6r}{r^2} \right) $$ $$ = 6(1 + r) \cdot \frac{6(1 + r)}{r^2} $$ $$ = \frac{36(1 + r)^2}{r^2} $$ Notice that this can be rewritten as a perfect square: $$ = \left( \frac{6(1 + r)}{r} \right)^2 = \left( \frac{6}{r} + \frac{6r}{r} \right)^2 $$ $$ = \left( \frac{6}{r} + 6 \right)^2 $$
Step 5: Substitute $a_2$ back into the expression and apply the outer square.
From Step 2, we know that $a_2 = \frac{6}{r}$. $$ \left( \frac{6}{r} + 6 \right)^2 = (a_2 + 6)^2 $$ The original question asks for the square of this entire factored value: $$ [(6 + a_2)^2]^2 = (6 + a_2)^4 $$