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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

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