Question
Let $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ be an A.P. of four terms such that each term of the A. P. and its common difference $l$ are integers. If $\alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} = 48$ and $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4} + l^{4} = 361$ then the largest term of the A.P. is equal to
Options
Solution
$a_{1},a_{2},a_{3},a_{4}$ as $a - 3\text{ }d,a - d,a + d,a + 3\text{ }d$
where $d = \frac{\mathcal{l}}{2}$
$${\because a_{1} + a_{2} + a_{3} + a_{4} = 48 \Rightarrow 4a = 48 \Rightarrow a = 12 }{\& a_{1}a_{2}a_{3}a_{4} + \mathcal{l}^{4} = 361 \Rightarrow \left( a^{2} - 9d^{2} \right)\left( a^{2} - d^{2} \right) + 16d^{4} }{= 361 }{\Rightarrow \left( 144 - 9{\text{ }d}^{2} \right)\left( 144 - d^{2} \right) + 16{\text{ }d}^{4} = 361 }{\Rightarrow 25{\text{ }d}^{4} - 1440{\text{ }d}^{2} + (144)^{2} = 361 }{\left( 5{\text{ }d}^{2} - 144 \right)^{2} = 19^{2} }$$$\therefore 5{\text{ }d}^{2} - 144 = 19$ or -19
$d^{2} = \frac{163}{5}$ or $d^{2} = \frac{125}{5} = 25$
$d = \sqrt{\frac{163}{5}}$ or $d = 5$
$\therefore\mathcal{l} = 2\sqrt{\frac{163}{5}}$ or $\mathcal{l} = 10$
(rejected)
∵ common difference is an integer
∴ largest term $= 12 + 15 = 27$
Create a free account to view solution
View Solution Free