In each of the following APs the 1st, 2nd and last terms are given. Find the common difference the number of terms in each sequence.

$$a_n = a_1 + (n-1)d$$ $$69 = 1 + (n-1)4$$ $$68 = (n-1)4$$ $$\frac{68}{4} = n-1$$ $$17 = n-1$$ $$n = 17 + 1$$ $$n = 18$$

The common difference is $\boxed{4}$ and the number of terms is $\boxed{18}$.

$$a_n = a_1 + (n-1)d$$ $$152 = 80 + (n-1)3$$ $$152 - 80 = (n-1)3$$ $$72 = (n-1)3$$ $$\frac{72}{3} = n-1$$ $$24 = n-1$$ $$n = 24 + 1$$ $$n = 25$$

The common difference is $\boxed{3}$ and the number of terms is $\boxed{25}$.

$$a_n = a_1 + (n-1)d$$ $$0 = 98 + (n-1)(-7)$$ $$-98 = (n-1)(-7)$$ $$\frac{-98}{-7} = n-1$$ $$14 = n-1$$ $$n = 14 + 1$$ $$n = 15$$

The common difference is $\boxed{-7}$ and the number of terms is $\boxed{15}$.

$$a_n = a_1 + (n-1)d$$ $$21 = 2.3 + (n-1)1.1$$ $$21 - 2.3 = (n-1)1.1$$ $$18.7 = (n-1)1.1$$ $$\frac{18.7}{1.1} = n-1$$ $$17 = n-1$$ $$n = 17 + 1$$ $$n = 18$$

The common difference is $\boxed{1.1}$ and the number of terms is $\boxed{18}$.

Convert $-2\frac{1}{2}$ to a decimal: $-2.5$.

$$d = a_2 - a_1 = -2.5 - (-6) = -2.5 + 6 = 3.5$$

$$a_n = a_1 + (n-1)d$$ $$71 = -6 + (n-1)3.5$$ $$71 + 6 = (n-1)3.5$$ $$77 = (n-1)3.5$$ $$\frac{77}{3.5} = n-1$$ $$22 = n-1$$ $$n = 22 + 1$$ $$n = 23$$

The common difference is $\boxed{3.5}$ and the number of terms is $\boxed{23}$.