The fourth term of an AP is one less than twice the second term. If the sixth term is 13, find (a) the first term (b) Common difference (c) the 16th term.

The fourth term ($T_4$) is one less than twice the second term ($T_2$):

$$T_4 = 2T_2 - 1$$

Substitute the formula for $T_n$:

$$a + (4-1)d = 2(a + (2-1)d) - 1$$ $$a + 3d = 2(a + d) - 1$$ $$a + 3d = 2a + 2d - 1$$

Rearrange to get the first equation:

$$a - d = 1 \quad (*)$$

The sixth term ($T_6$) is 13:

$$T_6 = 13$$

Substitute the formula for $T_n$:

$$a + (6-1)d = 13$$ $$a + 5d = 13 \quad (**)$$

From equation $(*)$, we can express $a$ in terms of $d$:

$$a = d + 1$$

Substitute this expression for $a$ into equation $(**)$:

$$(d + 1) + 5d = 13$$ $$6d + 1 = 13$$ $$6d = 13 - 1$$ $$6d = 12$$ $$d = \frac{12}{6}$$ $$d = 2$$

Now substitute $d=2$ back into $a = d+1$:

$$a = 2 + 1$$ $$a = 3$$

a) The first term
The first term is $a$.

$$a = 3$$

The first term is $\boxed{3}$.

b) Common difference
The common difference is $d$.

$$d = 2$$

The common difference is $\boxed{2}$.

c) The 16th term
To find the 16th term ($T_{16}$), use the formula $T_n = a + (n-1)d$ with $n=16$, $a=3$, and $d=2$:

$$T_{16} = a + (16-1)d$$ $$T_{16} = 3 + (15)(2)$$ $$T_{16} = 3 + 30$$ $$T_{16} = 33$$

The 16th term is $\boxed{33}$.