1. Let $1^{\text{st}}$ and $4^{\text{th}}$ terms be $x$ and $y$ respectively,

$1^{\text{st}} \text{difference} = 1-x; -7; y+6; -14-y; ..$

$2^{\text{nd}} \text{difference} = x-8; y+13; -2y-20; ...$

since the $2^{\text{nd}}$ difference is constant,

$x-8=y+13$ ... Equation 1

$y+13=-2y-20$ ... Equation 2

then from Equation 2,

\[3y=-33\]

\[ y= -11\]

and substitute in Equation 1,

\[x= -11+13+8= 10\]

hence the $2^{\text{nd}} \text{difference} = y+13= -11+13= 2$

2. $1^{\text{st}} \text{term} = x$

$= 10$ , calculated in question 1.