Ans: (a) $-3$ (b) $444$
- Let $d$ be the common difference.
$\begin{array}{rcl}
T(38) & = & T(1) + 37d \\
555 & = & 666 + 37d \\
37d & = & -111 \\
d & = & -3
\end{array}$ -
$\begin{array}{rcl}
S(n) & > & 0 \\
\dfrac{n}{2} [2(666) + (n-1)(-3)] & > & 0 \\
1335n -3n^2 & > & 0 \\
n^2 – 445n & < & 0 \\ n( n - 445) & < & 0 \end{array}$Therefore, $0 < n < 445$. Hence, the greatest value of $n$ is $444$.
