Which term of the $A.P.$ $85, 78, 71, \ldots$ is its first negative term? If the order of this term is $n$,find $S_{n}$.

(A) The given $A.P.$ is $85, 78, 71, \ldots$ where the first term $a = 85$ and the common difference $d = 78 - 85 = -7$.
For the $n^{th}$ term to be the first negative term,we set $a_{n} < 0$.
The formula for the $n^{th}$ term is $a_{n} = a + (n - 1)d$.
So,$85 + (n - 1)(-7) < 0$.
$85 - 7n + 7 < 0$.
$92 - 7n < 0$.
$7n > 92$.
$n > 13.14$.
Since $n$ must be an integer,the first negative term is the $14^{th}$ term $(n = 14)$.
To find $S_{14}$,we use the sum formula $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.
$S_{14} = \frac{14}{2}[2(85) + (14 - 1)(-7)]$.
$S_{14} = 7[170 + 13(-7)]$.
$S_{14} = 7[170 - 91]$.
$S_{14} = 7[79] = 553$.