p,q,r  are positive and are in A.P.

∴ q =  $\frac{p+r}{2}$              .........(i)

∵ The roots of p $x^2$ + qx + r = 0 are real

⇒$q^2$ ≥ 4pr ⇒  ${\left[\frac{p+r}{2}\right]}^2$ ≥ 4pr  [using (i)]

⇒$p^2$ + $r^2$ - 14pr ≥ 0

⇒ ${\left(\frac{r}{p}\right)}^2$ - 14 ($\frac{r}{p}$) + 1 ≥ 0      ( ∵ p>0 and p ≠ 0)

 ${(\frac{r}{p}-7)}^2$  -  $({4\sqrt{3})}^2$  ≥ 0  

⇒| $\frac{r}{p}$ - 7| ≥ 4 $\sqrt{3}$.

We know that A > G > H

Where A is arithmetic mean, G is geometric mean and H is harmonic mean, then A > G

$\Rightarrow \frac{a+b}{2}>\sqrt{ab}$ or (a + b) > 2$\sqrt{ab}$

$T_{\pi }$ =  $n^2$(n + 1) =  $n^3$ +  $n^2$

$S_{\pi }$ = $\sum$$T_{\pi }$ = $\sum$$n^3$ + $\sum$$n^2$ = ${\left[\frac{n(n+1)}{2}\right]}^2$ + $\frac{n(n+1)(2n+1)}{6}$ 

=  $\frac{n(n+1)}{2}$[$\frac{n(n+1)}{2}$ +  $\frac{2n+1}{3}$] =  $\frac{n(n+1)(3n^2+7n+2)}{12}$

Given that $\frac{a+ar+a{r}^2}{a+ar+a{r}^2+a{r}^3+a{r}^4+a{r}^5}=\frac{125}{152}.$

           $\Rightarrow \frac{1+r+r^2}{1+r+r^2+r^3+r^4+r^5}=\frac{125}{152}\text{i.e.,}\frac{1+r+r^2}{(1+r+r^2)+r^3(1+r+r^2)}=\frac{125}{152}\Rightarrow \frac{1}{1+r^3}=\frac{125}{152}\Rightarrow r^3=\frac{152}{125}-1=\frac{27}{125}\Rightarrow r=\frac{3}{5}$

The given condition can be expressed as 
$T_n$ = $T_{n+1}+T_{n+2},$ where $n\ge 1.$ 
If $a$ is the first term and $r$ is the common ratio, then by the condition is
$a{r}^{n-1}=a{r}^n+a{r}^{n+1}.\Rightarrow r^{n-1}=r^n(1+r)$
$\Rightarrow r^2+r-1=0$
$\Rightarrow r=\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}$

Since each term is positive, the common ratio of the GP is $\frac{\sqrt{5}-1}{2}.$ 

$T_{\pi }$ =  $n^2$ + n  ⇒ $S_{\pi }$ = $\sum T_{\pi }$ = $\sum$  $n^2$ + $\sum$n

=  $\frac{n(n+1)(2n+1)}{6}$ +  $\frac{n(n+1)}{2}$

=  $\frac{n(n+1)}{6}$ {2n + 1 + 3} =  $\frac{n(n+1)(n+2)}{3}$

Let $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k.$
$\Rightarrow a=k^x,b=k^y,c=k^z.$

Since $a,b,c$ are in G.P., $b^2=ac.$
$\therefore k^{2y}=k^x.k^z=k^{x+z}$
$\therefore 2y=x+z$
$\Rightarrow x,y,z$ are in A.P.

If a and d be the first term and common difference of the A.P.

Then $T_p$ = a + (p - 1)d, $T_q$ = a + (q - 1)d and

$T_r$ = a+(r - 1)d.

If $T_p,T_q,T_r$ are in G.P.

Then its common ratio R = $\frac{T_q}{T_p}$ =  $\frac{T_r}{T_q}$ =  $\frac{T_q-T_r}{T_p-T_q}$

= $\frac{[a+(q-1\left)d\right]-[a+(r-1\left)d\right]}{[a+(p-1\left)d\right]-[a+(q-1\left)d\right]}$ = $\frac{q-r}{p-q}$

Similarly, we can show that R =$\frac{q-r}{p-q}$ = $\frac{r-s}{q-r}$

Hence (p - q), (q - r),(r - s) be in G.P.

Given, $\frac{S_7}{S_{11}}=\frac{6}{11}and130<t_7<140$
$\Rightarrow \frac{\frac{7}{2}[2a+6d]}{\frac{11}{2}[2a+10d]}=\frac{6}{11}$
 $\Rightarrow \frac{7(2a+6d)}{(2a+10d)}=6$
$\Rightarrow a=9d\dots \dots \left(i\right)$
Also, $130<t_7<140$
$\Rightarrow 130<a+6d<140$
$\Rightarrow 130<9d+6d<140$   [from Eq. (i)]
$\Rightarrow 130<15d<140$
$\Rightarrow \frac{26}{3}<d<\frac{28}{3}$  
$\therefore$ d=9   [since, d is a natural number]