Question

Positive numbers a₁, a_2, .... a₅ are in geometric progression. Their mean and variance are \(\frac{31}{10}\) and \(\frac{m}{n}\) respectively. The mean of the reciprocals is \(\frac{31}{40}\), then m + n is?

• A. 209
• B. 211
• C. 113
• D. 429

Answer 1

The Correct Option is B

Let the terms of the sequence be: 

\( \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \)

Step 1: Given Equations

From the problem:

\( \frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 5 \times \frac{31}{10} \quad \dots (1) \)

And:

\( \frac{r^2}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} = 5 \times \frac{31}{40} \quad \dots (2) \)

Step 2: Divide (1) by (2)

Divide equation (1) by equation (2):

\( \frac{\frac{a^2}{r^2} + \frac{a^2}{r} + a^2 + a^2r + a^2r^2}{r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2} } = \frac{31}{10} \frac{40}{31} = 4 \)

From this:

\( a^2 = 4 \Rightarrow a = 2 \text{ (since } a > 0 \text{)} \)

Step 3: Solve for r

From \( r + \frac{1}{r} = \frac{5}{2} \) :

\( r = 2 \)

Step 4: Sequence Terms

The sequence terms are:

\( \frac{1}{2}, 1, 2, 4, 8 \)

Step 5: Variance ( \( \sigma^2 \) )

Using the formula for variance:

\( \sigma^2 = \frac{\Sigma x^2}{N} - \left(\frac{\Sigma x}{N}\right)^2 \)

Compute \( \Sigma x \) and \( \Sigma x^2 \):

\( \Sigma x = \frac{1}{2} + 1 + 2 + 4 + 8 = 15.5 \)

\( \Sigma x^2 = (\frac{1}{2})^2 + 1^2 + 2^2 + 4^2 + 8^2 = \frac{1}{4} + 1 + 4 + 16 + 64 = 85.25 \)

Substitute into the formula:

\( \sigma^2 = \frac{85.25}{5} - \left(\frac{15.5}{5}\right)^2 \)

Simplify:

\( \sigma^2 = \frac{186}{25} \)

From:

\( \frac{186}{25} = \frac{M}{N}, \quad M + N = 186 + 25 = 211 \)

Final Answer: \( M + N = 211 \)

Answer 2

The correct answer is option (B) : 211