Answers

2014-06-08T18:43:39+05:30
Suppose the 1st ,2nd , 3rd, 4th , 5th terms are x,y,z,a,b respectively
average of 5 terms is 18.4
so
[x+y+z+a+b]/5=18.4
so
x+y+z+a+b=92................................1
average of first three terms is 12
[x+y+z]/3=12
so
x+y+z=36...................................2
average of last three terms is 30
so
[z+a+b]/3=30
so
z+a+b=90.....................................3
substituting the value of eq 2 in eq 1
a+b=92-36=56.........................4
substituting the value of a+b in eq 3
z=90-56=34
so the value of third term is 34
1 5 1
2014-06-08T20:09:40+05:30
Given the average of 5 numbers is 18.4
Let us consider those 5 numbers be a,b,c,d,e
The average of 5 numbers = 18.4
(a+b+c+d+e)/5 = 18.4
(a+b+c+d+e) = 18.4 * 5 = 92 --------------------> 1
The average of first three numbers is = 12
(a+b+c)/3 = 12
(a+b+c) = 12*3 = 36 ----------------------> 2
The average of last three numbers is = 30
(c+d+e)/3 = 30
(c+d+e) = 90 ------------------------------> 3
By adding 2 and 3 we get 
a+b+c+c+d+e = 36+90 = 126
(a+b+c+d+e)+c = 126 
but from it is clear (a+b+c+d+e) = 92
By substituting this in above we get
92+c = 126 
c = 126 - 92 = 34

1 5 1
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