\(\sqrt { { x }^{ 2 }+25 } -\sqrt { { x }^{ 2 }+9 } =x\left( 1+\frac { 25 }{ { x }^{ 2 } } \right) ^{ 1/2 }-x\left( 1+\frac { 9 }{ { x }^{ 2 } } \right) ^{ 1/2 }\)
\(=x\left[ 1+\frac { 1 }{ 2 } \left( \frac { 25 }{ { x }^{ 2 } } \right) +\frac { \frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } \right) }{ 1.2 } \left( \frac { 25 }{ { x }^{ 2 } } \right) ^{ 2 }+.... \right] -x\left[ 1+\frac { 1 }{ 2 } \left( \frac { 9 }{ { x }^{ 2 } } \right) +\frac { \frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } \right) }{ 1.2 } \left( \frac { 9 }{ { x }^{ 2 } } \right) ^{ 2 }+.... \right] \)
\(=x+\frac { 25 }{ 2x } -\frac { 625 }{ 8{ x }^{ 3 } } +.....-x-\frac { 9 }{ 2x } +\frac { 81 }{ 8{ x }^{ 3 } } +....=\frac { 16 }{ 2x } =\frac { 8 }{ x } approximately\) 

Here, \(\frac{S_{3}}{S_{6}}=\frac{125}{152}\)
⇒ \(\frac{a(r^{3}-1)/(r-1)}{a(r^{6}-1)/(r-1)}=\frac{125}{152}\Rightarrow \frac{r^{3}-1}{r^{6}-1}=\frac{125}{152}\)
∴ \(\frac{r^{3}-1}{(r^{3}-1)(r^{3}+1)}=\frac{125}{152}\Rightarrow \frac{1}{r^{3}+1}=\frac{125}{152}\)
∴ 152 = 125 r³ + 125 or 125r³ = 27
ஃ r3=\(\frac{27}{125}=(\frac{3}{5})^{3}\)
⇒ r = {\((\frac{3}{5})^{3}\)}^1/3 = \(\frac{3}{5}\)
Hence, the common ratio of the G.P. is \(\frac{3}{5}\)

Let A be the first term and D, the common difference of A.P.
a_p = a, ∴ A + (p -1)D = a --- (1)
aq = b, ∴ A+(q-1)D = b --- (2)
a, = c, ∴ A+(r-1)D = c --- (3)
∴ a (q - r) + b (r - p) + c (p - q) = [A + (p - 1) D] (q - r) + [A + (q -1) D]
(r - p) + [A + (r - 1) D] (p - q) [Using (1), (2) and (3)]
= (q - r + r - p + P - q) A + [ (p -1)(q - r) + (q -1)(r - p) + (r -1) (p - q)] D
= (0) A + (pq - pr - q + r + qr - pq - r + p + pr - p - qr + q) D
= (0) A + (0) D =0.

Here T₂-T₁ = \(log\frac { { a }^{ 2 } }{ b } -log\quad a=log\frac { { a }^{ 2 }/b }{ a } \)
= \(log\frac { a }{ b } \)
T₃-T₂ = \(log\frac { a^{ 3 } }{ b^{ 2 } } -log\frac { a^{ 2 } }{ b^{ 2 } } =log\frac { a^{ 3 } }{ b^{ 2 } } +log\frac { a^{ 2 } }{ b^{ 2 } } \)
= \(log\frac { a^{ 3 } }{ b^{ 2 } } \times \frac { b }{ { a }^{ 2 } } =log\frac { a }{ b } \)
∴ T₂-T₁ = T₃-T₂ = \(log\left( \frac { a }{ b } \right) \)
∴ The given sequences is an A.P

Using binomial theorem, we have
(1 + ax)ⁿ-1 + nC₁(ax) + nC₂(ax)²+.......+aⁿxⁿ
= \(1+nax+{{n(n-1)}\over{2}}a^2x^2+....a^nx^n\)
Given (1 + ax)ⁿ = 1 + 12x + 64x² +....
Conparing the Co-efficient of x and x², we get
n a = 12
and \({n(n-1)\over 2}a^2=64\)
\((n-1).{na.a\over2}=64\Rightarrow(n-1){(12)a\over2}=64\)
\((n-1)6a=64\Rightarrow(n-1)a={{64}\over{6}}\)    \(\left[ \because na=12\Rightarrow a={12\over n} \right]\)
\(\Rightarrow(n-1)\left( {12\over n} \right)={64 \over 6}\)
\(={n-1\over n}={ 64 \over 6\times 12}\Rightarrow{n-1\over n}={8\over 9}\)
\(\Rightarrow\)  9n - 9 = 8n
\(\Rightarrow\)  n = 9 and \(a=\frac { 12 }{ n } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)