Question

Let \( a_1, a_2, a_3, \dots \) be a G.P. of increasing positive terms. If \( a_1 a_5 = 28 \) and \( a_2 + a_4 = 29 \), then the value of \( a_6 \) is equal to:

• A. 628
• B. 526
• C. 784
• D. 812

Hint:

In geometric progressions, use the common ratio \(r\) and apply the equations involving terms to find the missing terms.

Answer 1

The Correct Option is C

Solution to the Geometric Progression Problem 

We are given a geometric progression (G.P.) where:

• \( a_1 a_5 = 28 \)
• \( a_2 + a_4 = 29 \)

Step 1: Express the Terms of the G.P. in Terms of \( a \) and \( r \)

In a geometric progression, the terms are given by: - \( a_1 = a \) - \( a_2 = a r \) - \( a_3 = a r^2 \) - \( a_4 = a r^3 \) - \( a_5 = a r^4 \) - \( a_6 = a r^5 \) Now, let's use the given conditions:

Step 2: Use the Given Conditions

Condition 1: \( a_1 a_5 = 28 \)

\[ a \cdot a r^4 = 28 \quad \Rightarrow \quad a^2 r^4 = 28 \quad \text{...(1)} \]

Condition 2: \( a_2 + a_4 = 29 \)

\[ a r + a r^3 = 29 \quad \Rightarrow \quad a r (1 + r^2) = 29 \] \[ \Rightarrow a^2 r^2 (1 + r^2)^2 = 29^2 \quad \text{...(2)} \]

Step 3: Solve the System of Equations

From equation (1): \[ a^2 r^4 = 28 \] From equation (2): \[ a^2 r^2 (1 + r^2)^2 = 29^2 \] Dividing equation (2) by equation (1): \[ \frac{r^2}{(1 + r^2)^2} = \frac{28}{29^2} \] \[ \Rightarrow \frac{r}{1 + r^2} = \frac{\sqrt{28}}{29} \] Therefore: \[ r = \sqrt{28} \]

Step 4: Solve for \( a \)

Now substitute \( r = \sqrt{28} \) into equation (1): \[ a^2 r^4 = 28 \] \[ a^2 \times (28)^2 = 28 \] \[ a^2 \times 784 = 28 \quad \Rightarrow \quad a^2 = \frac{28}{784} = \frac{1}{28} \] \[ a = \frac{1}{\sqrt{28}} \]

Step 5: Calculate \( a_6 \)

Now, we calculate \( a_6 \): \[ a_6 = a r^5 \] Substituting \( a = \frac{1}{\sqrt{28}} \) and \( r = \sqrt{28} \): \[ a_6 = \frac{1}{\sqrt{28}} \times (28)^2 \times \sqrt{28} \] \[ a_6 = \frac{1}{\sqrt{28}} \times 784 \times \sqrt{28} = 784 \]

Conclusion

Therefore, the value of \( a_6 \) is: \[ \boxed{784} \]