Question

If \( a_1, a_2, a_3, \ldots \) are in increasing geometric progression such that \[ a_1 + a_3 + a_5 = 21, a_1 a_3 a_5 = 64, \] then \( a_1 + a_2 + a_3 \) is:

• A. \(5\)
• B. \(7\)
• C. \(10\)
• D. \(15\)

Hint:

For GP problems: \begin{itemize} \item Express all required terms using \( a \) and \( r \) \item Use substitution (like \( r^2 = x \)) to simplify equations \item Check conditions such as increasing/decreasing GP \end{itemize}

Answer 1

The Correct Option is B

Step 1: Let the GP be: \[ a_1 = a, a_2 = ar, a_3 = ar^2, a_4 = ar^3, a_5 = ar^4 \] Since the GP is increasing, \( r>1 \).
Step 2: Using the given conditions: \[ a_1 + a_3 + a_5 = a(1 + r^2 + r^4) = 21 (1) \] \[ a_1 a_3 a_5 = a^3 r^6 = 64 \] \[ \Rightarrow (ar^2)^3 = 64 \Rightarrow ar^2 = 4 (2) \]  
Step 3: From (2), \( a = \dfrac{4}{r^2} \). Substitute in (1): \[ \frac{4}{r^2}(1 + r^2 + r^4) = 21 \] Let \( r^2 = x \,(>1) \): \[ \frac{4}{x}(1 + x + x^2) = 21 \] \[ 4x^2 - 17x + 4 = 0 \] \[ x = \frac{17 \pm 15}{8} \Rightarrow x = 4 \;(\text{since } x>1) \] 
Step 4: Hence, \[ r^2 = 4 \Rightarrow r = 2, a = \frac{4}{4} = 1 \] 
Step 5: Required sum: \[ a_1 + a_2 + a_3 = 1 + 2 + 4 = 7 \]