If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms E in the sequence are smaller than the average? Choose the correct choice:
A) 5
B) 6
C) 7
D) 8
E) 9
Let the sequence have N terms starting 18*K   --&gt; 18K, 18(K+1), 18(K+2), .... , 18(K+N-1) is the sequence.   The sum of the sequence is   ---&gt; 18K + 18(K+1) + 18(K+2) + .... + 18(K+N-1) =18*(K+(K+1)+...+(K+N-1)) = 18*(N*(2K+N-1)/2).   So the average is   153=[18*(N*(2K+N-1)/2)]/N = 18*(2K+N-1)/2 = 9 *(2K+N-1) ---&gt; by cancelling 9 we get 2K+N-118*(2 + -1)/2= 
<gadget id="calculator">18*(2 + -1)/2</gadget>
<output>9</output>
9 *(2K+N-1) ---&gt; by cancelling 9 we get 2K+N-1=17 --&gt; 2K+N=18 ---*).   On the other hand, since the greatest term is 270, 18(K+N-1)=270 --&gt; by cancelling 18 we get K+N-1=15 --&gt; K+N=16 ---**).   By *), **) we have K=2, N=14.   Since 153/18= 
<gadget id="calculator">153/18</gadget>
<output>17/2 = around 8.5</output>
8.5, 18*8&lt;153&lt;18*9. So the number of terms E which are smaller than the average is 18*2, 18*3, ..... ,18*8 ---&gt; 7. The answer is 7.
<result>C</result>
If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms E in the sequence are smaller than the average?