Question 854209: There are four numbers in an Arithmetic Progression. The sum of the two extremes is 8, and the product of the middle two numbers is 15. What are the numbers?
Found 2 solutions by aditya2117, richwmiller:
Answer by aditya2117(32)   (Show Source):
You can put this solution on YOUR website! Let the first term be a and the C.D be d.
The series is a,a+d,a+2d,a+3d
Accordingly,
a + a + 3d = 8 ......................................... (1)
(a+d)(a+2d)= 15......................................... (2)
from (1),
a= 8-3d/2
Now in (2),
(8-3d/2 + d)(8-3d/2 + 2d) = 15
=> (8-3d+2d)/2*(8-3d+4d)/2 = 15
=> (8-d)(8+d) = 15*4
=> 64-d^2=60
=> d^2 = 64-60 =4
=> d = +2 or -2
a = 8-3.2/2
= 8-6/2
=1
a = 8-3.(-2)/2
= 8+6/2
=14/2=7
The series is : (1,3,5,7) / (7,5,3,1)
Both are same. (ANS.)
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! an = a + (n - 1)d
a1+a4 = 2a+ 3d=8
a2*a3=15
(a + 1d)( a + 2d)=15
a^2+ad+2ad+2d^2=15
a^2+3ad+2d^2=15,
2a+ 3d=8
Two possible progressions
a = 1, d = 2
1,3,5,7
or
a = 7, d = -2
7,5,3,1
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