(A) How many terms are there in an A.P. whose first and fifth terms are - 14 and 2, respectively and the last term is 62.

\[\begin{aligned} & a=-14, a_{5}=2 \quad \Rightarrow a+4 d=2 \\ & \quad-14+4 d=2 \Rightarrow d=4 \\ & a_{n}=62 \quad \Rightarrow a+(n-1) d=62 \\ & -14+(n-1) 4=62 \quad \Rightarrow n=20 \end{aligned} \]

$\displaystyle 65$, $\displaystyle 61$, $\displaystyle 57$, 53...

\[a=65, d=-4 \]

Let \(\displaystyle \mathrm{a}_{\mathrm{n}}\) be the first negative term

\[\begin{aligned} & \mathrm{a}_{\mathrm{n}}<0 \Rightarrow \mathrm{a}+(\mathrm{n}-1) \mathrm{d}<0 \\ & 65+(\mathrm{n}-1)(-4)<0 \Rightarrow 69-4 \mathrm{n}<0 \\ & \quad \mathrm{n}>\frac{69}{4} \end{aligned} \]

∴ Least positive integral value of n which satisfies n > \(\displaystyle \frac{69}{4}\) is $\displaystyle 18$

\(\displaystyle \therefore 1^{\text {st }}\) negative term of the \(\displaystyle \mathrm{AP}=18\)