Question 1121963
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The sum of the first 100 terms is 15000, so the average of the terms is 15000/100 = 150.<br>
In particular, the average of the first and last terms is 150; and the sum of the 50th and 51st terms is 150.<br>
The sub-sequence has 33 terms with a sum of 5016, so the average is 5016/33 = 152.<br>
In particular, the 17th (middle) term alone is 152.<br>
The 17th term of the sub-sequence is the 51st term of the full sequence.<br>
So we know the average of the 50th and 51st terms is 150; and we know that the 51st term is 152.  That means the 50th term is 148.<br>
With the 50th term 148 and the 51st term 152, the common difference is 4.<br>
With a common difference of 4 and 51st term 152, the first term is 152-50(4) = 152-200 = -48.<br>
Answer: first term -48; common difference 4.<br>
CHECK:
a1+a2+...+a100 = (-48)+(-44)+...+(348) = 100((-48)+348)/2 = 100(150) = 15000
a3+a6+...+a99 = (-40)+(-28)+...+344) = 33((-40)+344)/2 = 33(152) = 5016