Question

Sum of first three terms of a GP is 93 and the product of the first and third terms is 225. Then the 4th term of the GP is

• A. 75
• B. 375
• C. 225
• D. 525

Hint:

For sum and product of GP terms, always look for the common ratio \( r \) by substituting \( a = \text{constant}/r \).

Answer 1

The Correct Option is B

Concept: Let the first three terms of the Geometric Progression (GP) be \( a, ar, ar^2 \).

Step 1: Set up equations based on given information.
Sum of first three terms: \[ a + ar + ar^2 = 93 \implies a(1 + r + r^2) = 93 \quad \cdots (1) \] Product of the first and third terms: \[ a \times ar^2 = 225 \implies (ar)^2 = 225 \implies ar = 15 \implies a = \frac{15}{r} \quad \cdots (2) \]

Step 2: Substitute \( a \) into the sum equation.
\[ \frac{15}{r}(1 + r + r^2) = 93 \] \[ 15\left(\frac{1}{r} + 1 + r\right) = 93 \] \[ \frac{1}{r} + r + 1 = \frac{93}{15} = 6.2 \] \[ r + \frac{1}{r} = 5.2 \implies r^2 - 5.2r + 1 = 0 \] Multiply by 5 to clear decimals: \( 5r^2 - 26r + 5 = 0 \).

Step 3: Solve for \( r \).
\[ (5r - 1)(r - 5) = 0 \implies r = 5 \text{ or } r = \frac{1}{5} \] If \( r = 5 \),
then \( a = 15/5 = 3 \).
The GP is \( 3, 15, 75, \dots \).
The 4th term = \( ar^3 = 3 \times 5^3 = 3 \times 125 = 375 \).