The first three terms of an arithmetic progression (AP) are p/6, 2p, 6 and p respectively. (a) Given that the common difference of the AP is not zero, find the value of p. (b) Find the 20th term of the AP. (c) Calculate the sum of the first 20 terms of the AP.

The first three terms of the AP are given as $T_1 = \frac{p^2}{6}$, $T_2 = 2p-6$, and $T_3 = p$.

In an AP, the common difference ($d$) is constant, meaning $T_2 - T_1 = T_3 - T_2$.
Substitute the given terms into this equation:

Multiply the entire equation by 6 to eliminate the fraction:

Rearrange the terms to form a quadratic equation:

Multiply by -1 to make the leading coefficient positive:

We need two numbers that multiply to 72 and add up to -18. These numbers are -6 and -12.

This gives two possible values for $p$:

If $p=6$:
$T_1 = \frac{6^2}{6} = 6$
$T_2 = 2(6)-6 = 12-6 = 6$
$T_3 = 6$
The terms are 6, 6, 6. The common difference $d = 6-6 = 0$. This contradicts the condition that the common difference is not zero.

If $p=12$:
$T_1 = \frac{12^2}{6} = \frac{144}{6} = 24$
$T_2 = 2(12)-6 = 24-6 = 18$
$T_3 = 12$
The terms are 24, 18, 12. The common difference $d = 18-24 = -6$. This is not zero.
Therefore, the value of $p$ is 12.

First term, $a = T_1 = 24$.
Common difference, $d = T_2 - T_1 = 18 - 24 = -6$.

We need to find the 20th term, so $n=20$.

We need to calculate the sum of the first 20 terms, so $n=20$. We have $a=24$ and $T_{20}=-90$.