sum of three no.s in a.p is -3 and their product is 8 .find the no.s
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Heya !!!
Let the required Numbers are ( A - D) , A , ( A + D)
According to question,
Sum of three numbers = -3
A- D + A + A + D = -3
3A = -3
A = -3/3
A = -1
And,
Product of three numbers = 8
(A-D) (A) (A+D) = 8
A (A² - D²) = 8
=> -1 { (-1)² - (D)² } = 8
=> -1 + D² = 8
=> D² = 8+1
=> D = ✓9
=> D = 3
Thus,
A = -1 and D = 3
Required Numbers are :
( A - D) , ( A ) , ( A + D)
( -1 + 3) , -1 , ( -1 + 3)
2 , -1 , 2
HOPE IT WILL HELP YOU....... :-)
Let the required Numbers are ( A - D) , A , ( A + D)
According to question,
Sum of three numbers = -3
A- D + A + A + D = -3
3A = -3
A = -3/3
A = -1
And,
Product of three numbers = 8
(A-D) (A) (A+D) = 8
A (A² - D²) = 8
=> -1 { (-1)² - (D)² } = 8
=> -1 + D² = 8
=> D² = 8+1
=> D = ✓9
=> D = 3
Thus,
A = -1 and D = 3
Required Numbers are :
( A - D) , ( A ) , ( A + D)
( -1 + 3) , -1 , ( -1 + 3)
2 , -1 , 2
HOPE IT WILL HELP YOU....... :-)
Honey9927:
plz help me again there is one more ques
if theta1,theta2,theta3,.......
thetan are in a.p then sec theta1×sec theta2 + sec theta2×sec theta3+...........+sec theta(n-1)×sec theta n =tan theta n-tan /sin×d
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