Question

Let $ a_1, a_2, a_3, ... $ be a G.P. of increasing positive numbers. If $ a_3 a_5 = 729 $ and $ a_2 + a_4 = \frac{111}{4} $, then $ 24(a_1 + a_2 + a_3) $ is equal to

• A. 131
• B. 130
• C. 129
• D. 128

Hint:

Use the given conditions to form equations and solve for the first term and common ratio of the G.P.

Answer 1

The Correct Option is C

To solve this problem, we first need to understand and apply the properties of a geometric progression (G.P.). In a G.P., each term is obtained by multiplying the previous term by a fixed, non-zero number known as the common ratio, \( r \).

Given that \( a_1, a_2, a_3, \ldots \) is a G.P. of positive numbers, the terms are as follows:

• \( a_1 = a_1 \) (first term)
• \( a_2 = a_1 \cdot r \)
• \( a_3 = a_1 \cdot r^2 \)
• \( a_4 = a_1 \cdot r^3 \)
• \( a_5 = a_1 \cdot r^4 \)

From the problem, we have two key pieces of information:

1. \( a_3 \cdot a_5 = 729 \)
2. \( a_2 + a_4 = \frac{111}{4} \)

We'll use these pieces of information to find \( a_1 + a_2 + a_3 \) and finally compute \( 24(a_1 + a_2 + a_3) \).

Step 1: Use the first piece of information:

\(a_3 \cdot a_5 = (a_1 \cdot r^2) \cdot (a_1 \cdot r^4) = a_1^2 \cdot r^6 = 729\)

Therefore, we can write:

\((a_1 \cdot r^3)^2 = 729 \implies a_1 \cdot r^3 = \sqrt{729} = 27\)

Step 2: Use the second piece of information:

\(a_2 + a_4 = a_1 \cdot r + a_1 \cdot r^3 = a_1 \cdot (r + r^3) = \frac{111}{4}\)

Step 3: Find the relation between \( r \) using the values obtained:

From Step 1, \(a_1 \cdot r^3 = 27\). Let's use it in the equation from Step 2:

\(\frac{a_1 \cdot r^3}{r^2} + a_1 \cdot r^3 = \frac{111}{4}\)

Let \( x = a_1 \cdot r^3 \), hence:

\(\frac{x}{r^2} + x = \frac{111}{4}\)

With \( x = 27 \):

\(\frac{27}{r^2} + 27 = \frac{111}{4}\)

Solving this equation:

\(\frac{27}{r^2} = \frac{111}{4} - 27 = \frac{3}{4}\)

\(\Rightarrow 27 \cdot 4 = 3 \cdot r^2 \Rightarrow r^2 = \frac{108}{3} = 36 \Rightarrow r = 6\)

Hence, the common ratio \( r = 6 \), and from the equation \( a_1 \cdot r^3 = 27 \), we have:

\(a_1 \cdot 216 = 27 \Rightarrow a_1 = \frac{27}{216} = \frac{1}{8}\)

Step 4: Now, calculate \( a_1 + a_2 + a_3 \):

\[\begin{align*} a_1 &= \frac{1}{8} \\ a_2 &= \frac{1}{8} \cdot 6 = \frac{3}{4} \\ a_3 &= \frac{1}{8} \cdot 6^2 = \frac{9}{2} \end{align*}\]

Sum, \( a_1 + a_2 + a_3 = \frac{1}{8} + \frac{3}{4} + \frac{9}{2} = \frac{1}{8} + \frac{6}{8} + \frac{36}{8} = \frac{43}{8} \).

Finally, compute \( 24(a_1 + a_2 + a_3) \):

\(24 \cdot \frac{43}{8} = 3 \cdot 43 = 129\)

Thus, the value of \( 24(a_1 + a_2 + a_3) \) is 129.

Answer 2

Let the \( 1^{st} \) term of G.P. be a and common ratio be r.
Given
\( a_3 a_5 = 729 \)
\( ar^2 \cdot ar^4 = 729 \)
\( a^2 r^6 = 729 \)
\( ar^3 = 27 \) ... (i)
Also, \( a_2 + a_4 = \frac{111}{4} \)
\( ar + ar^3 = \frac{111}{4} \)
\( ar = \frac{111}{4} - 27 = \frac{111 - 108}{4} = \frac{3}{4} \) ... (ii)
Dividing (i) by (ii):
\( \frac{ar^3}{ar} = \frac{27}{3/4} \)
\( r^2 = 36 \)
\( r = 6 \) (since the G.P. is increasing, r>0) From (ii):
\( a(6) = \frac{3}{4} \)
\( a = \frac{1}{8} \) Now, \( 24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) \)
\( = 24a(1 + r + r^2) \)
\( = 24 \cdot \frac{1}{8} (1 + 6 + 36) \)
\( = 3(43) \)
\( = 129 \)