Four different integers form an increasing AP. One of these numbers is equal to the sum of the squares of the other three numbers. Then,
(i) The smallest number is-
Text Solution
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Ans.
(i)
Sol. Let the four integer be a – d, a, a + d and a + 2d where a and b are integers and d > 0.
∴ a + 2d = (a – d) 2 + a 2 + (a + d) 2 ⇒ 2d
2 – 2d + 3a 2 – a = 0 ….(i)
∴ d = 
…(ii)
Since d is positive integer
∴ 1 + 2a – 6a 2 > 0
⇒ 6a 2 – 2a – 1 < 0
⇒ 
< a < 
[ ∴ a is an integer ]
∴ a = 0, put in Eq. (ii), we get d = 1 or 0 but d > 0
∴ d = 1
∴ The four numbers are : –1, 0, 1, 2.
(ii)
Sol. Let the four integer be a – d, a, a + d and a + 2d where a and b are integers and d > 0.
∴ a + 2d = (a – d) 2 + a 2 + (a + d) 2 ⇒ 2d
2 – 2d + 3a 2 – a = 0 ….(i)
∴ d = 
…(ii)
Since d is positive integer
∴ 1 + 2a – 6a 2 > 0
⇒ 6a 2 – 2a – 1 < 0
⇒ 
< a < 
[ ∴ a is an integer ]
∴ a = 0, put in Eq. (ii), we get d = 1 or 0 but d > 0
∴ d = 1
∴ The four numbers are : –1, 0, 1, 2.
(iii)
Sol. Let the four integer be a – d, a, a + d and a + 2d where a and b are integers and d > 0.
∴ a + 2d = (a – d) 2 + a 2 + (a + d) 2 ⇒ 2d
2 – 2d + 3a 2 – a = 0 ….(i)
∴ d = 
…(ii)
Since d is positive integer
∴ 1 + 2a – 6a 2 > 0
⇒ 6a 2 – 2a – 1 < 0
⇒ 
< a < 
[ ∴ a is an integer ]
∴ a = 0, put in Eq. (ii), we get d = 1 or 0 but d > 0
∴ d = 1
∴ The four numbers are : –1, 0, 1, 2.
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