The arithmetic mean of the three numbers a, b, c is 8.
Find the arithmetic mean of the four numbers: a + 1, b + 2, c + 6, 3.
A. 7
B. 11
C. 5
D. 9
E. 27
The arithmetic mean A is defined by the formula:
A = \frac{1}{n}\sum_{i=1}^{n}a_i = \frac{a_1 + a_2 + \ldots + a_n}{n}
So the arithmetic mean of three numbers a, b, c is A_1 =\frac{a+b+c}{3} = 8.
The arithmetic mean for four numbers a+1, b+2, c+6, 3 is A_2 =\frac{(a+1)+(b+2)+(c+6)+3}{4} = \frac{a+b+c+9+3}{4}= \frac{a+b+c+12}{4} = \frac{a+b+c}{4} +\frac{12}{4}
= \frac{a+b+c}{4} + 3.
From the first arithmetic mean, we can express what a+b+c is and use it in the second mean in order to get the value needed.
A_1=\frac{a+b+c}{3} = 8 \to a+b+c = 8 \cdot 3\to a+b+c = 24.
Using this information in the A_2 we get:
A_2 = \frac{a+b+c}{4} + 3 = \frac{24}{4} + 3 = 6 + 3 = 9.
Which gives us the correct solution (D) 9.
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