in winter, the temperature at a hill station from monday to friday is in A.P .the sum of the temperature of monday ,tuesday and wednesday is zero degree celsius and the sum of the temperature of thursday and friday is 15 degree celcius .find the temperature of each of the five days
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a-2d monday -1
a-d tuesday -2
a wednesday -3
a+d thursday -4
a+2d friday -5
adding eqn -1 -2 -3 -4 & -5
we get
a-2d+a-d+a+a+d+a+2d=0+15 °c
5a=15
a=3
from eqn 1 2 &3
a-2d+a-d+a=0
3a-3d=0
a=d
a=d=3
monday -3°c
tuesday 0° c
wednesday 3° c
thursday 6°c
friday 9°c
a-2d monday -1
a-d tuesday -2
a wednesday -3
a+d thursday -4
a+2d friday -5
adding eqn -1 -2 -3 -4 & -5
we get
a-2d+a-d+a+a+d+a+2d=0+15 °c
5a=15
a=3
from eqn 1 2 &3
a-2d+a-d+a=0
3a-3d=0
a=d
a=d=3
monday -3°c
tuesday 0° c
wednesday 3° c
thursday 6°c
friday 9°c
saurabh125:
thank u
Answered by
22
Answer:
Step-by-step explanation:
Let the five days are,
a-2d , a-d , a , a+d , a+2d
Mon, Tue, wed, Thu , Fri
Acc to first condition,
a-2d + a-d + a = 0
3a - 3d = 0 --------( 1 )
Acc to second condition,
a+d + a+2d = 15
2a + 3d = 15 --------( 2 )
Adding ( 1 ) and ( 2 ),
3a - 3d = 0
+ 2a + 3d = 15
____________
5a = 15
a = 15/5
a = 3
Put a=3 in ( 1 )
3a - 3d = 0
3×3 - 3d =0
9 - 3d = 0
-3d =0-9
d = 9/3
d = 3
a-2d = 3-2×3= 3-6 = -3
a-d = 3-3= 0
a= 3
a+d = 3 +3 = 6
a+2d= 3+ 2×3 = 3+6 = 9
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