the sum of three numbers in AP is –3 and the product is 8 . find the numbers
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Answered by
5
Let the three numbers be a-d, a, a+d
Given that,
a-d + a + a+d = - 3
3a = - 3
=> a = - 1........(1)
Also given that,
(a-d) *a*(a+d) = 8
(a^2 - d^2)*a = 8
Putting a = - 1,
(-1 - d^2)(-1) = 8
(1 + d^2) = 8
d^2 = 7
=> d = +- (7)^1/2
So the numbers are -1 - (7)^1/2 , -1 , -1 + (7)^1/2
Given that,
a-d + a + a+d = - 3
3a = - 3
=> a = - 1........(1)
Also given that,
(a-d) *a*(a+d) = 8
(a^2 - d^2)*a = 8
Putting a = - 1,
(-1 - d^2)(-1) = 8
(1 + d^2) = 8
d^2 = 7
=> d = +- (7)^1/2
So the numbers are -1 - (7)^1/2 , -1 , -1 + (7)^1/2
mahekgupta:
not beneficial
Based on the information provided in the question, this is the only solution possible.
In case there is an error, kindly check the information provided.
I got the error.
d will be either +3 or - 3
So numbers will be 2, - 1, - 4
Answered by
4
Let the numbers (a - d), a & (a + d) are in AP
a/q,
(a - d) + a + (a + d) = -3 --------------- (i)
& (a - d) x (a) x (a + d) = 8 --------------- (ii)
from (i)
=> 3a = -3 => a = -1
from(ii)
=> (-1 - d)(-1)(-1 + d) = 8 (Substitute the value of a = -1 in (ii))
=> (d + 1)(d - 1) = 8
=> d² - 1² = 8
=> d² = 8 + 1 = 9
=> d = ± √9
=> either d = +3 or -3
take d = 3 and a = -1, Numbers will be: (-4, -1, 2)
take d = -3 and a = -1, Numbers will be: (2, -1, -4)
a/q,
(a - d) + a + (a + d) = -3 --------------- (i)
& (a - d) x (a) x (a + d) = 8 --------------- (ii)
from (i)
=> 3a = -3 => a = -1
from(ii)
=> (-1 - d)(-1)(-1 + d) = 8 (Substitute the value of a = -1 in (ii))
=> (d + 1)(d - 1) = 8
=> d² - 1² = 8
=> d² = 8 + 1 = 9
=> d = ± √9
=> either d = +3 or -3
take d = 3 and a = -1, Numbers will be: (-4, -1, 2)
take d = -3 and a = -1, Numbers will be: (2, -1, -4)
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