find four numbers in G. P. Such that their product is 64 and sum of the second and third number is 6
Answers
Answer:
1,2,4,8
Step-by-step explanation:
Let the numbers be a,b,c,d
By problem ,
b+ c= 6
=> c = 6 - b
SINCE the number are in GP,
so , we can say that ,
=> b ² = ac
=> a = (b²)/c
Also , c² = bd ,
=> d = (c²)/b
BY PROBLEM,
=> a*b*c*d = 64
=> {(b²)/c}*b*c*{(c²)/b} = 64
=> b²c²=64
=> bc = √64 = 8
=> b(6-b)=8
=> 6b - b² = 8
=> b² - 6b +8 =0
=> (b-4)(b-2)=0
=> b= 2
so, c =6-2 = 4
=> a = b²/c = 4/4 = 1
=> d = c²/b= 14/2 = 8
so , the numbers are = 1,2,4,8
Answer:
1,2,4,8
Step-by-step explanation:
Let the numbers be a,b,c,d
By problem ,
b+ c= 6
=> c = 6 - b
SINCE the number are in GP,
so , we can say that ,
=> b ² = ac
=> a = (b²)/c
Also , c² = bd ,
=> d = (c²)/b
BY PROBLEM,
=> a*b*c*d = 64
=> {(b²)/c}*b*c*{(c²)/b} = 64
=> b²c²=64
=> bc = √64 = 8
=> b(6-b)=8
=> 6b - b² = 8
=> b² - 6b +8 =0
=> (b-4)(b-2)=0
=> b= 2
so, c =6-2 = 4
=> a = b²/c = 4/4 = 1
=> d = c²/b= 14/2 = 8
so , the numbers are = 1,2,4,8