Question

The two geometric means between 2 and 8 are $g_{1}$ and $g_{2}$. Then $g_{1}^{3}+g_{2}^{3}$ is equal to

• A. 160
• B. 140
• C. 180
• D. 164
• E. 174

Hint:

Math Tip: You don't need to find the exact decimal value of $r$ (which is $\sqrt[3]{4}$). By writing the required expression in terms of $r^3$, you can directly substitute the integer value to save time.

Answer 1

The Correct Option is A

Concept:
Sequences and Series - Insertion of Geometric Means.
Step 1: Set up the geometric progression.
When two geometric means, $g_1$ and $g_2$, are inserted between 2 and 8, the sequence becomes a Geometric Progression (G.P.): $$ 2, g_1, g_2, 8 $$ Here, the first term $a = 2$, and the fourth term $t_4 = 8$.
Step 2: Find the common ratio (r).
The formula for the $n$-th term of a G.P. is $t_n = a \cdot r^{n-1}$.
For the fourth term: $$ t_4 = a \cdot r^{4-1} $$ $$ 8 = 2 \cdot r^3 $$ $$ r^3 = 4 $$
Step 3: Express the geometric means in terms of the common ratio.
The geometric means can be written as the second and third terms of the sequence: $$ g_1 = a \cdot r = 2r $$ $$ g_2 = a \cdot r^2 = 2r^2 $$
Step 4: Calculate the cubes of the geometric means.
We need to find the value of $g_1^3 + g_2^3$. First, cube each term: $$ g_1^3 = (2r)^3 = 8r^3 $$ $$ g_2^3 = (2r^2)^3 = 8r^6 = 8(r^3)^2 $$
Step 5: Substitute the value of $r^3$ and solve.
Substitute $r^3 = 4$ into the expressions from Step 4: $$ g_1^3 = 8(4) = 32 $$ $$ g_2^3 = 8(4)^2 = 8(16) = 128 $$ Now, add them together: $$ g_1^3 + g_2^3 = 32 + 128 = 160 $$