It is given that the 3rd term and the 19th term of an arithmetic sequence are −55 and 17 respectively.

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Uniteasy Admin · Accepted Answer

Using the formula for general terms of an arithmetic sequence
$$a_3=a_1+2d=-55$$
$$a_{19}=a_1+18d= 17$$
Solving the function system yields
$$d= \dfrac{9}{2}  $$
$$a_1 = -64$$
Therefore, the general term
 $$T_n=a_1+(n-1)d$$
$$=-64+(n-1)\dfrac{9}{2}  $$
$$=\dfrac{9}{2}n-\dfrac{137}{2}  $$ If $$T_n <0$$
$$\dfrac{9}{2}n-\dfrac{137}{2} < 0$$
$$n<\dfrac{137}{9}  = 15\dfrac{2}{5}  $$
So when $$n = 15$$, the sequence has the greatest negative term 
$$T_{15} = \dfrac{9}{2}	imes 15 -\dfrac{137}{2}=-1$$
So the greatest negative integral term of the sequence is -1

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