The sum of three consecutive terms in an AP is 6 and their product is -120.
Find the three numbers.
Answers
Answer:
The three consecutive terms of the AP are either - 6, 2 & 10 or 10, 2 & - 6.
Step-by-step-explanation:
Let the first term of the AP be "a" and common difference be "d".
The three consecutive terms of the AP are a, a + d & a + 2d.
From the first condition,
a + ( a + d ) + ( a + 2d ) = 6
⇒ a + a + d + a + 2d = 6
⇒ a + a + a + d + 2d = 6
⇒ 3a + 3d = 6
⇒ a + d = 2 - - - [ Dividing both sides by 3 ]
⇒ a = 2 - d
⇒ a = - d + 2 - - - ( 1 )
From the second condition,
a ( a + d ) * ( a + 2d ) = - 120
⇒ ( - d + 2 ) * [ ( - d + 2 ) + d ] * [ ( - d + 2 ) + 2d ] = - 120 - - - [ From ( 1 ) ]
⇒ ( - d + 2 ) * ( - d + 2 + d ) * ( - d + 2 + 2d ) = - 120
⇒ ( - d + 2 ) * ( - d + d + 2 ) * ( - d + 2d + 2 ) = - 120
⇒ ( - d + 2 ) * ( 0 + 2 ) * ( d + 2 ) = - 120
⇒ ( - d + 2 ) * 2 * ( d + 2 ) = - 120
⇒ ( - d + 2 ) * ( d + 2 ) * 2 = - 120
⇒ [ ( 2 )² - ( - d)² ] * 2 = - 120 - - - [ ∵ ( a + b ) ( a - b ) = a² - b² ]
⇒ ( 4 - d² ) * 2 = - 120
⇒ 4 - d² = - 120 ÷ 2
⇒ - d² + 4 = - 60
⇒ - d² = - 60 - 4
⇒ - d² = - 64
⇒ d² = 64
⇒ d = ± 8
Now, by substituting d = 8 in equation ( 1 ), we get,
a = - d + 2 - - - ( 1 )
⇒ a = - 8 + 2
⇒ a = - 6
Now,
a + d = - 6 + 8
⇒ a + d = 2
Now,
a + 2d = - 6 + 2 * 8
⇒ a + 2d = - 6 + 16
⇒ a + 2d = 10
Now, by substituting d = - 8 in the equation ( 1 ), we get,
a = - d + 2 - - - ( 1 )
⇒ a = - ( - 8 ) + 2
⇒ a = 8 + 2
⇒ a = 10
Now,
a + d = 10 + ( - 8 )
⇒ a + d = 10 - 8
⇒ a + d = 2
Now,
a + 2d = 10 + 2 * ( - 8 )
⇒ a + 2d = 10 - 16
⇒ a + 2d = - 6
∴ The three consecutive terms of the AP are either - 6, 2 & 10 or 10, 2 & - 6.
The three consecutive terms of the AP are either - 6, 2 & 10 or 10, 2 & - 6.
Step-by-step-explanation:
Let the first term of the AP be "a" and common difference be "d".
The three consecutive terms of the AP are a, a + d & a + 2d.
From the first condition,
a + ( a + d ) + ( a + 2d ) = 6
⇒ a + a + d + a + 2d = 6
⇒ a + a + a + d + 2d = 6
⇒ 3a + 3d = 6
⇒ a + d = 2 - - - [ Dividing both sides by 3 ]
⇒ a = 2 - d
⇒ a = - d + 2 - - - ( 1 )
From the second condition,
a ( a + d ) * ( a + 2d ) = - 120
⇒ ( - d + 2 ) * [ ( - d + 2 ) + d ] * [ ( - d + 2 ) + 2d ] = - 120 - - - [ From ( 1 ) ]
⇒ ( - d + 2 ) * ( - d + 2 + d ) * ( - d + 2 + 2d ) = - 120
⇒ ( - d + 2 ) * ( - d + d + 2 ) * ( - d + 2d + 2 ) = - 120
⇒ ( - d + 2 ) * ( 0 + 2 ) * ( d + 2 ) = - 120
⇒ ( - d + 2 ) * 2 * ( d + 2 ) = - 120
⇒ ( - d + 2 ) * ( d + 2 ) * 2 = - 120
⇒ [ ( 2 )² - ( - d)² ] * 2 = - 120 - - - [ ∵ ( a + b ) ( a - b ) = a² - b² ]
⇒ ( 4 - d² ) * 2 = - 120
⇒ 4 - d² = - 120 ÷ 2
⇒ - d² + 4 = - 60
⇒ - d² = - 60 - 4
⇒ - d² = - 64
⇒ d² = 64
⇒ d = ± 8
Now, by substituting d = 8 in equation ( 1 ), we get,
a = - d + 2 - - - ( 1 )
⇒ a = - 8 + 2
⇒ a = - 6
Now,
a + d = - 6 + 8
⇒ a + d = 2
Now,
a + 2d = - 6 + 2 * 8
⇒ a + 2d = - 6 + 16
⇒ a + 2d = 10
Now, by substituting d = - 8 in the equation ( 1 ), we get,
a = - d + 2 - - - ( 1 )
⇒ a = - ( - 8 ) + 2
⇒ a = 8 + 2
⇒ a = 10
Now,
a + d = 10 + ( - 8 )
⇒ a + d = 10 - 8
⇒ a + d = 2
Now,
a + 2d = 10 + 2 * ( - 8 )
⇒ a + 2d = 10 - 16
⇒ a + 2d = - 6
∴ The three consecutive terms of the AP are either - 6, 2 & 10 or 10, 2 & - 6.