Question

Let a₁, a₂, a₃, … be a G.P of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then a₁ a₉+a₂ a₄ a₉+a₅+a₇ is equal to __________.

Answer 1

Correct Answer: 60

Step 1: General form of terms in a geometric progression. In a geometric progression (GP), the general term is given by: \[ a_n = a_1 r^{n-1}, \] where \( a_1 \) is the first term and \( r \) is the common ratio. 

Step 2: Using the given conditions. We are given the product of the fourth and sixth terms: \[ a_4 \cdot a_6 = 9. \] Using the general form of the terms: \[ a_4 = a_1 r^3 \quad \text{and} \quad a_6 = a_1 r^5, \] so: \[ a_1 r^3 \cdot a_1 r^5 = 9 \quad \Rightarrow \quad a_1^2 r^8 = 9. \quad \cdots (1) \] We are also given the sum of the fifth and seventh terms: \[ a_5 + a_7 = 24. \] Using the general form of the terms: \[ a_5 = a_1 r^4 \quad \text{and} \quad a_7 = a_1 r^6, \] so: \[ a_1 r^4 + a_1 r^6 = 24 \quad \Rightarrow \quad a_1 r^4(1 + r^2) = 24. \quad \cdots (2) \] 

Step 3: Solving equations (1) and (2). From equation (1): \[ a_1^2 r^8 = 9 \quad \Rightarrow \quad a_1^2 = \frac{9}{r^8}. \quad \cdots (3) \] From equation (2): \[ a_1 r^4 (1 + r^2) = 24 \quad \Rightarrow \quad a_1 = \frac{24}{r^4(1 + r^2)}. \quad \cdots (4) \] Substitute equation (4) into equation (3): \[ \left( \frac{24}{r^4(1 + r^2)} \right)^2 r^8 = 9 \quad \Rightarrow \quad \frac{576 r^8}{r^8 (1 + r^2)^2} = 9 \quad \Rightarrow \quad \frac{576}{(1 + r^2)^2} = 9. \] Solving this: \[ (1 + r^2)^2 = \frac{576}{9} = 64 \quad \Rightarrow \quad 1 + r^2 = 8 \quad \Rightarrow \quad r^2 = 7. \] 

Step 4: Finding \( a_1 \) and solving for the desired expression. From equation (3), substitute \( r^2 = 7 \) into the equation: \[ a_1^2 = \frac{9}{r^8} = \frac{9}{(7)^4} = \frac{9}{2401} \quad \Rightarrow \quad a_1 = \frac{3}{49}. \] Now, we compute the expression \( a_1a_9 + a_2a_8 + a_3a_7 + a_4a_6 \). The terms are: \[ a_9 = a_1 r^8, \quad a_8 = a_1 r^7, \quad a_7 = a_1 r^6, \quad a_6 = a_1 r^5, \] so: \[ a_1a_9 = a_1^2 r^8, \quad a_2a_8 = a_1^2 r^6, \quad a_3a_7 = a_1^2 r^4, \quad a_4a_6 = a_1^2 r^8. \] Summing these terms: \[ a_1a_9 + a_2a_8 + a_3a_7 + a_4a_6 = a_1^2 (r^8 + r^6 + r^4 + r^8). \] Simplifying: \[ = a_1^2 (2r^8 + r^6 + r^4). \] Substitute \( r^2 = 7 \): \[ r^4 = 49, \quad r^6 = 343, \quad r^8 = 2401. \] Thus: \[ a_1a_9 + a_2a_8 + a_3a_7 + a_4a_6 = a_1^2 (2 \times 2401 + 343 + 49) = a_1^2 (4802 + 343 + 49) = a_1^2 \times 5194. \] Finally: \[ a_1a_9 + a_2a_8 + a_3a_7 + a_4a_6 = \frac{9}{2401} \times 5194 = 60. \]