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

To solve this problem, we need to first establish the terms of the geometric progression (GP). Given that the sequence \( a_1, a_2, a_3, \ldots \) is in an increasing geometric progression, we can express the terms as:

• \( a_1 = a \)
• \( a_2 = ar \)
• \( a_3 = ar^2 \)
• \( a_4 = ar^3 \)
• \( a_5 = ar^4 \)

From the problem statement, we have two equations:

1. \( a_1 + a_3 + a_5 = 21 \), which translates to \( a + ar^2 + ar^4 = 21 \).
2. \( a_1 a_3 a_5 = 64 \), which means \( a \cdot ar^2 \cdot ar^4 = 64 \).

Let's simplify the second equation first:

\( a \cdot ar^2 \cdot ar^4 = a^3 r^{2+4} = a^3 r^6 = 64 \)

This results in:\( (ar^2)^3 = 64 \)

Taking the cube root on both sides, we get:

\( ar^2 = 4 \)

Now substituting \( ar^2 = 4 \) back into the first equation:

\( a + 4 + \frac{4}{r^2} = 21 \)

Thus, the equation becomes: \(\frac{4}{r^2} = 21 - a - 4 = 17 - a\)

So, we now have:

\( ar^2 = 4 \)

and

\( \frac{4}{r^2} = 17 - a \). \)

Solving these equations simultaneously:

Substitute \( r^2 = \frac{4}{a} \) in \( \frac{4}{r^2} = 17 - a \):

\( \frac{4}{\frac{4}{a}} = 17 - a \)

\( a = 17 - a \)

\( a = \frac{9}{2} \) (solving for \( a \)).

Now, find \( a_1 + a_2 + a_3 \):

\( a_1 + a_2 + a_3 = a + ar + ar^2 \)

= \( a + a (\sqrt{2}) + ar^2 \)

= \( \frac{9}{2} + \frac{9}{2}(\sqrt{2}) + \frac{9}{2}\)

= \( 9 + \frac{9}{2}\)

= \( 9 = 18/2\)

= \(7\).