Question

If \( a_1, a_2, a_3, \dots \) are the terms of an increasing geometric progression such that \[ a_1 + a_3 + a_5 = 21, \quad a_1a_3a_5 = 64 \] then \[ a_1 + a_2 + a_3 \] is

• A. 5
• B. 7
• C. 10
• D. 15

Hint:

For geometric progressions, always use the ratio of terms to express terms in terms of a single variable and simplify accordingly.

Answer 1

The Correct Option is B

To solve this problem, we must first understand the properties of a geometric progression. Given is a sequence \(a_1, a_2, a_3, \dots\) which is an increasing geometric progression.

We know two conditions:

1. \(a_1 + a_3 + a_5 = 21\)
2. \(a_1a_3a_5 = 64\)

For a geometric progression, each term can be expressed in terms of the first term \(a_1\) and the common ratio \(r\). Thus:

• \(a_2 = a_1 r\)
• \(a_3 = a_1 r^2\)
• \(a_5 = a_1 r^4\)

Now, substituting these into the equation:

1. \(a_1 + a_3 + a_5 = 21\)
   • \(a_1 + a_1 r^2 + a_1 r^4 = 21\)
   • \(a_1 (1 + r^2 + r^4) = 21\)
2. \(a_1 a_3 a_5 = 64\)
   • \(a_1 \cdot a_1 r^2 \cdot a_1 r^4 = 64\)
   • \(a_1^3 r^6 = 64\)

From the equation \(a_1^3 r^6 = 64\), we can solve for \(a_1\) and \(r\):

• \(a_1^3 (r^6) = 64\)
• Taking the cube root of both sides: \(a_1 r^2 = 4\)

Substitute \(r^2 = \frac{4}{a_1}\) into the equation \(a_1(1 + r^2 + r^4) = 21\):

1. \(1 + \frac{4}{a_1} + \frac{16}{a_1^2} = \frac{21}{a_1}\)
2. Multiplying through by \(a_1^2\) to eliminate the fractions gives:
3. \(a_1^2 + 4a_1 + 16 = 21a_1\)
4. Rearrange to form a quadratic equation:
5. \(a_1^2 - 17a_1 + 16 = 0\)

Use the quadratic formula to solve for \(a_1\):

\(a_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -17\), and \(c = 16\).

• \(a_1 = \frac{17 \pm \sqrt{289 - 64}}{2}\)
• \(a_1 = \frac{17 \pm \sqrt{225}}{2}\)
• \(a_1 = \frac{17 \pm 15}{2}\)

The two potential solutions for \(a_1\) are:

• \(a_1 = \frac{32}{2} = 16\)
• \(a_1 = \frac{2}{2} = 1\)

Since \((a_1 = 16)\) would make the sequence constant (which contradicts being increasing), we select \(a_1 = 1\). With \(a_1 = 1\), and we already derived \(a_1r^2 = 4\); so \(r^2 = 4\), giving \(r = 2\).

Now calculate \(a_1 + a_2 + a_3\):

• \(a_1 = 1\), \(a_2 = 2\), \(a_3 = 4\).
• Thus, \(a_1 + a_2 + a_3 = 1 + 2 + 4 = 7\).

Therefore, the sum \(a_1 + a_2 + a_3\) is 7.