Question

Suppose a₁, a₂, 2, a₃, a4 be in an arithmeticogeometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is \(\frac{49}{2}\) , then a4 is equal to ________ .

Answer 1

Correct Answer: 16

The given terms are \( \frac{a - 2d}{4}, \frac{a - d}{2}, a, 2(a + d), 4(a + 2d) \). 
Given that \( a = 2 \), we substitute this into the terms: \[ \frac{a - 2d}{4}, \frac{a - d}{2}, a, 2(a + d), 4(a + 2d) \Rightarrow \frac{2 - 2d}{4}, \frac{2 - d}{2}, 2, 2(2 + d), 4(2 + 2d) \] Now, use the given sum of the 5 terms being \( \frac{49}{2} \): \[ \left( \frac{1}{4} + \frac{1}{2} + 1 + 6 \right) \times 2 + (-1 + 2 + 8)d = \frac{49}{2} \] \[ 2 \left( \frac{3}{4} + 7 \right) + 9d = \frac{49}{2} \] \[ 2 \times \frac{31}{4} + 9d = \frac{49}{2} \] \[ \frac{62}{4} + 9d = \frac{49}{2} \] Now solve for \( d \): \[ 9d = \frac{49}{2} - \frac{62}{4} = \frac{98}{4} - \frac{62}{4} = \frac{36}{4} = 9 \] Thus, \( d = 1 \). Now, substitute \( d = 1 \) into \( a_4 = 4(a + 2d) \): \[ a_4 = 4(2 + 2 \times 1) = 4(2 + 2) = 4 \times 4 = 16 \] Thus, the value of \( a_4 \) is \( 16 \).