Let be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If and then the largest term of the A.P. is equal to [2026]

Question

Let $a_{1}, a_{2}, a_{3}, a_{4}$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_{1} + a_{2} + a_{3} + a_{4} = 48$ and $a_{1} a_{2} a_{3} a_{4} + l^{4} = 361,$ then the largest term of the A.P. is equal to [2026]

Smart Achievers · Accepted Answer

(4)

$a_{1}, a_{2}, a_{3}, a_{4} as a - 3 d, a - d, a + d, a + 3 d$

$where d = \frac{ℓ}{2}$

$∵$ $a_{1} + a_{2} + a_{3} + a_{4} = 48 \Rightarrow 4 a = 48 \Rightarrow a = 12$

$& a_{1} a_{2} a_{3} a_{4} + ℓ^{4} = 361 \Rightarrow (a^{2} - 9 d^{2}) (a^{2} - d^{2}) + 16 d^{4} = 361$

$\Rightarrow (144 - 9 d^{2}) (144 - d^{2}) + 16 d^{4} = 361$

$\Rightarrow 25 d^{4} - 1440 d^{2} + {(144)}^{2} = 361$

${(5 d^{2} - 144)}^{2} = 19^{2}$

$∴$ $5 d^{2} - 144 = 19 or - 19$

$d^{2} = \frac{163}{5} or d^{2} = \frac{125}{5} = 25$

$d = \sqrt{\frac{163}{5}} or d = 5$

$∴$ $ℓ = 2 \sqrt{\frac{163}{5}} or ℓ = 10$

$(rejected)$

$∵$ common difference is an integer

$∴$ $largest term = 12 + 15 = 27$

Solution

Q.
Let $a_{1}, a_{2}, a_{3}, a_{4}$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_{1} + a_{2} + a_{3} + a_{4} = 48$ and $a_{1} a_{2} a_{3} a_{4} + l^{4} = 361,$ then the largest term of the A.P. is equal to [2026]

1 23

2 21

3 24

4 27

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