the sum of three numbers of AP is 3 and the product of the first and third number is -35.find the three numbers
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s3 = 3
3/2 ( 2a + (3-1)d ) = 3
3 ( 2a + (2)(d) ) = 6
2a + 2d = 2
a + d = 1 , a = (1-d). eq.1
a2 = 1
( a ) ( a + 2d ) = -35
a^2 + 2ad = -35
( 1 - d )^2 + 2 ( 1 - d )(d) = -35
1 + d^2 - 2d + 2d - 2d^2 = -35
1 - d^2 = -35
36 = d^2
d = 6 or -6
Putting 6 in eq.1
a + 6 = 1
a = -5
d = 6
The terms are -5 , 1 , 7
Putting -6 in eq.1
a - 6 = 1
a = 7
d = -6
The terms are 7 , 1 , -5
The three no.s are 7 , 1 , - 5.
These can be arranged in both orders , increasing as well as decreasing.
3/2 ( 2a + (3-1)d ) = 3
3 ( 2a + (2)(d) ) = 6
2a + 2d = 2
a + d = 1 , a = (1-d). eq.1
a2 = 1
( a ) ( a + 2d ) = -35
a^2 + 2ad = -35
( 1 - d )^2 + 2 ( 1 - d )(d) = -35
1 + d^2 - 2d + 2d - 2d^2 = -35
1 - d^2 = -35
36 = d^2
d = 6 or -6
Putting 6 in eq.1
a + 6 = 1
a = -5
d = 6
The terms are -5 , 1 , 7
Putting -6 in eq.1
a - 6 = 1
a = 7
d = -6
The terms are 7 , 1 , -5
The three no.s are 7 , 1 , - 5.
These can be arranged in both orders , increasing as well as decreasing.
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Answer:
the numbers are -5, 1, 7 or 7, 1 ,-5
see the attachment for the solution.
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