Let a be the first term and r the common ratio of the G.P.
∴ S= a + ar + ar²+ ...+ ar^{n - 1}
\(=\frac{a(1-r^{n})}{1-r}\)  --- (1)
p =\(a\times ar \times ar^{2}\times...\times ar^{n-1}=a^{n}r^{1+2+3}+..+(n-1)=a^{n}r^{n{(n-1})/2}\)
∴ \(P^{2}=a^{2n}r^{n(n-1)}\)   --- (2)
\(R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^{2}}+....+\frac{1}{ar^{n-1}}\)
⇒ \(R=\frac{1}{a}.\frac{(1-\frac{1}{r^{n}})}{(1-\frac{1}{r})}=\frac{(r^{n}-1)}{(r-1)}.\frac{1}{ar^{n-1}}\)   [∵Here, r<1]
∴ \(\frac{S}{R}=a\frac{(1-r^{n})}{1-r}.\frac{r-1}{r^{n}-1} ar^{n-1}= a^{2}r ^{n-1}\)
∴ \((\frac{S}{R})^{n}=a^{2n}r^{n(n-1)}\)   --- (3)
From (2) and (3) we get \(P^{2}=(\frac{S}{R})^{n}\)

Suppose a² = x and \(\sqrt{a^{2}-1}=y\)
∴ \(a^{2}+\sqrt{a+1}=x+y \) and \(a^{2}-\sqrt{a^{2}-1}=x-y\).
Using binomial theorem,
\((x+y)^{4}={^4C_{0}x^{4}+{^4C_{2}x^{3}y}+{^4C_{2}x^{2}y^{2}}+{^4C_{3}xy^{3}+{^4C_{4}y^{4}}}}\)
⇒ (x+y)⁴ = x⁴+ 4x³y + 6x²y² + 4xy³ + y⁴ and
(x-y)⁴ = ⁴C₀x⁴ - ⁴C₁x³y + ⁴C₂x²y² - ⁴C₃xy³ + ⁴C₄y⁴
⇒ \((x-y)^{4}=x^{4}-4x^{3}y+6x^{2}y^{2}-4xy^{3}+y^{4}\)
∴ \((x+y)^{4}+(x-y)^{4}=2[x^{4}+6x^{2}y^{2}+y^{4}]\)
\(\therefore (a^{2}+\sqrt{a^{2}+1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}=2[(a^{2})^{4}+6(a^{2})^{2}(\sqrt{a^{2}-1})^{2}+(\sqrt{a^{2}-1})^{4}] \)
= \(2[a^{8}+6a^{4}(a^{2}-1)+(a^{2}-1)^{2}]\)
= \(2[a^{8}+6a^{6}-6a^{4}+a^{4}-2a^{2}+1]\)
= \(2[a^{8}+6a^{6}-5a^{4}-2a^{2}+1]\)

Since \(\alpha ,\beta \) are the roots of the equation \(​​{ x }^{ 2 }-px+q=0\),we have
\(\alpha +\beta =-\frac { \left( -p \right) }{ 1 } =p\quad and\quad \alpha \beta =\frac { q }{ 1 } q\)
\(\therefore \log { (1+px+q{ x }^{ 2 }) } =\log { \left[ 1+(\alpha +\beta )x+\alpha \beta { x }^{ 2 } \right] } \)
\(=\log { \left[ (1+\alpha x)(1+\beta x) \right] } \)
\(=\log { (1+\alpha x)+log(1+\beta x) } \)
\(=\left( \alpha x-\frac { { \alpha }^{ 2 } }{ 2 } { x }^{ 2 }+\frac { { \alpha }^{ 3 } }{ 2 } { x }^{ 3 }+......+\infty \right) +\left( \beta x-\frac { { \beta }^{ 2 }{ x }^{ 2 } }{ 2 } +\frac { { \beta }^{ 3 }{ x }^{ 3 } }{ 3 } +.....\infty \right) \)
\(=(\alpha +\beta )x-\frac { { (\alpha }^{ 2 }+\beta ^{ 2 }) }{ 2 } { x }^{ 2 }+\frac { { (\alpha }^{ 3 }+\beta ^{ 3 }) }{ 3 } { x }^{ 3 }-...\infty \)
Hence proved. 

Let the positive numbers be a, b. Let a > b.
\(\therefore A=\frac { a+b }{ 2 } \Rightarrow a+b=2A \ \text {and} G=\sqrt { ab } \Rightarrow ab\Rightarrow { G }^{ 2 }\)
we know, (a + b)² = (a-b)² + 4ab
\(\Rightarrow\) 4A² = (a - b)² + 4G²
\(\Rightarrow\) (a - b)² = 4 (A² -G²)
 \(\Rightarrow\) a-b = 2\(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
(1)+(3) \(\rightarrow\)2a = 2\(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
\(\Rightarrow\) )a = A+ \(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
(1)-( 3) \(\rightarrow\) 2b = 2\(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
\(\Rightarrow\) b = A -\(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
The Numbers are A\(\pm \) \(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)

\(\frac { { \left( 1-2x \right) }^{ \frac { 2 }{ 3 } }{ \left( 4+5x \right) }^{ \frac { 3 }{ 2 } } }{ \sqrt { 1-x } } =\frac { \left( 1+\frac { 2 }{ 3 } (-2x)+..... \right) { \left( 4 \right) }^{ \frac { 3 }{ 2 } }{ \left( 1+\frac { 5 }{ 4 } x \right) }^{ \frac { 3 }{ 2 } } }{ { \left( 1-x \right) }^{ \frac { 1 }{ 2 } } } \)  [using binomial theorem for rational index]
\(={ \left( 1-\frac { 4x }{ 3 } \right) (8) }{ \left( 1+\frac { 3 }{ 2 } \left( \frac { 5 }{ 4 } x \right) \right) }\left( 1-\frac { 1 }{ 2 } (-x) \right) \) [neglecting x², x³ terms....]
\(=8\left( 1-\frac { 4x }{ 3 } \right) \left( 1+\frac { 15x }{ 8 } \right) \left( 1+\frac { x }{ 2 } \right) \)
\(=8\left( 1-\frac { 4x }{ 3 } +\frac { 15x }{ 8 } \right) \left( 1+\frac { x }{ 2 } \right) \)
\(=8\left( 1+\frac { 13x }{ 24 } \right) \left( 1+\frac { x }{ 2 } \right) =8\left( 1+\frac { 13x }{ 24 } +\frac { x }{ 2 } \right) =8\left( 1+\frac { 25x }{ 24 } \right) \)
When x = 0.001, the value of \(\frac { { \left( 1-2x \right) }^{ \frac { 2 }{ 3 } }{ \left( 4+5x \right) }^{ \frac { 3 }{ 2 } } }{ \sqrt { 1-x } } \)
= 8 + \(\frac { 25 }{ 3 } \)(0.001) = 8.01 (upto 2 places )