Question

find the four numbers in gp such that their product is 64 and the sum of the second and third number is 6​

Answer 1

Step-by-step explanation:

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