Seven numbers $$ a_1, a_2, a_3, a_4, a_5, a_6$$ and $$a_7$$ form an arithmetic sequence, where $$a_1

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Seven numbers $$ a_1, a_2, a_3, a_4, a_5, a_6$$ and $$a_7$$ form an arithmetic sequence, where $$a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7$$. It is given that the mean of the seven numbers is 0.

Uniteasy Admin · Accepted Answer

Let d be the common difference of the sequence
$$a_1=a_4-3d$$
$$a_2 = a_4-2d$$
$$a_3 = a_4-d$$
$$a_5=a_4+d$$
$$a_6=a_4+2d$$
$$a_7 = a_4+3d$$
$$a_1+a_2+\dots+a_7$$
$$=7a_4=0$$
Therefore, 
$$a_4 = 0$$ If $$a_4$$ is deleted, previous $$a_5$$ becomes the 4th term, previous $$a_6$$ becomes the 5th term, previous $$a_7$$ becomes 6th term.
$$a_1=a_4-3d$$
$$a_2 = a_4-2d$$
$$a_3 = a_4-d$$
$$a_4=a_4$$
$$a_5=a_4+d$$
$$a_6 = a_4+2d$$
$$a_1+a_2+\dots+a_6$$
$$=6a_4-3d=0$$
Therefore,
$$a_4=\dfrac{1}{2}d  $$
which means the numbers vary with the common difference.

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