Question

If a, b, c are in A. P. and a + 1, b, c + 3 are in G. P., arithmetic mean of a, b, c is 8, then the value of cube of geometric mean of a, b, c is:

• A. 312
• B. 314
• C. 318
• D. 128

Answer 1

The Correct Option is A

To solve the given problem, we start by analyzing the conditions for both Arithmetic Progression (A.P.) and Geometric Progression (G.P.).

1. Arithmetic Progression (A.P.) Condition:
   The terms \( a \), \( b \), and \( c \) are in an arithmetic progression, meaning they satisfy the relation: b = \frac{a + c}{2}
2. Geometric Progression (G.P.) Condition:
   The terms \( a + 1 \), \( b \), and \( c + 3 \) are in a geometric progression, satisfying the relation: b^2 = (a+1)(c+3)
3. Arithmetic Mean Condition:
   The arithmetic mean of \( a \), \( b \), and \( c \) is 8: \frac{a + b + c}{3} = 8 Simplifying gives: a + b + c = 24
4. Solving the System of Equations:
   1. From the A.P. condition: b = \frac{a + c}{2}.
   2. Using the arithmetic mean: a + b + c = 24.
   3. From the G.P. condition: b^2 = (a+1)(c+3).
5. Calculate Each Term:
   1. Substitute b = \frac{a + c}{2} into the arithmetic mean: a + \frac{a+c}{2} + c = 24
      Simplifying gives: 2a + a + c + 2c = 48
      3a + 3c = 48 or a + c = 16.
   2. Using a + c = 16 in the G.P condition: b^2 = (a + 1)(c + 3)
      Substitute c = 16 - a: b^2 = (a + 1)((16-a) + 3) = (a + 1)(19 - a)
   3. Since b = \frac{a + c}{2} = \frac{16}{2} = 8: 64 = (a+1)(19-a)
   4. Expanding gives: 64 = 19a - a^2 + 19 - a, simplifying: a^2 - 18a + 45 = 0.
6. Solving the Quadratic Equation:
   Solve a^2 - 18a + 45 = 0 using factorization or quadratic formula. a - 3 = 0 or a - 15 = 0.
   Thus, a = 3 and a = 15.
7. Substituting to find b and c:
   • For a = 3:
     c = 16 - 3 = 13
     So: b = \frac{3 + 13}{2} = 8
   • For a = 15:
     c = 16 - 15 = 1
     So: b = \frac{15 + 1}{2} = 8
8. Geometric Mean:
   The geometric mean of \( a \), \( b \), and \( c \) is: G = \sqrt[3]{abc}
   Checking each scenario:
   • For \( a = 3 \), \( b = 8 \), \( c = 13 \):
     G = \sqrt[3]{3 \cdot 8 \cdot 13} = \sqrt[3]{312}
     Thus, \( G^3 = 312 \), which is one of the options.
   • For \( a = 15 \), \( b = 8 \), \( c = 1 \):
     G = \sqrt[3]{15 \cdot 8 \cdot 1} = \sqrt[3]{120}, which doesn't match an option correctly.

Conclusion: The value of the cube of the geometric mean of \( a \), \( b \), and \( c \) is 312.