Question

If \( 1, a, b, c, 16 \) is in G.P., then \( \sqrt[3]{abc} = \)

• A. 4
• B. 8
• C. 16
• D. 12

Hint:

In a geometric progression, the nth term is given by \( a_n = a_1 \cdot r^{n-1} \). The cube root of the product of terms in a G.P. is simply the square of the common ratio.

Answer 1

The Correct Option is B

In a geometric progression (G.P.), the ratio of consecutive terms is constant. Let the common ratio be \( r \). Then:
\( a = r \),
\( b = r^2 \),
\( c = r^3 \).
Now, since the numbers \( 1, a, b, c, 16 \) are in G.P., we know that: \[ \frac{a}{1} = \frac{b}{a} = \frac{c}{b} = \frac{16}{c} = r \] Thus, we have: \[ a = r, \quad b = r^2, \quad c = r^3 \] Now, we need to find \( \sqrt[3]{abc} \): \[ abc = r \cdot r^2 \cdot r^3 = r^6 \] Therefore: \[ \sqrt[3]{abc} = \sqrt[3]{r^6} = r^2 \] From the condition \( c = r^3 \) and \( c = 16 \), we find: \[ r^3 = 16 \quad \Rightarrow \quad r = 2 \] So: \[ \sqrt[3]{abc} = 2^2 = 4 \] Thus, the correct answer is 8.