Let a be the first term and r be the common ratio of G.P.
∴ \(S_{1}=\frac{a(r^{n}-1)}{r-1}, S_{2}=\frac{a(r^{2n}-1)}{r-1}, S_{3}=\frac{a(r^{3n}-1)}{r-1}\)
where r ≠ 1
\(S_{3}-S_{2}= \frac{a}{r-1}(r^{3n}-r^{2n})=\frac{a(r^{n}-1)}{r-1}r^{2n}\)
\(S_{1}(S_{3}-S_{2})\frac{a(r^{n}-1)}{r-1}\times \frac{a(r^{n}-1)}{r-1}r^{2n}\)
=\([\frac{a(r^{n}-1)}{r-1}.r^{n}]^{2}\)   --- (1)
\((S_{2}-S_{1})=\frac{a}{r-1}(r^{2n}-r^{n})=\frac{a(r^{n}-1)}{r-1}r^{n}\)   --- (2)
∴ \(S_{1}(S_{3}-S_{2})=(S_{2}-S_{1})^{2}\) [From (1) and (2)]
When r = 1, S₁ = na, S₂ = 2na and S3 = 3 na
Then, \((S_{2}-S_{1})^{2}=2(na-na)^{2}=n^{2}a^{2}\) and \(S_{1}(S_{3}-S_{2})=na(3na-2na)\)
= na(na) = n² a²
∴ S₁ (S₃ - S₂) = (S₂ - S₁)² 

In the expansion of  (1+x)²ⁿ, Number of terms = 2n + 1, which is odd
There is only one middle term, the \(\frac{(2n+1)+1}{2}\) 1^th i.e. (n + 1)^th term
∴ T_n+1 is the only middle term T_n+1 = ²ⁿC_nxⁿ
Coefficient of T_n+1 = ²ⁿC_n  --- (1)
In the expansion of (1 + x)^2n-1,
Number of terms = 2n - 1 +1== 2n, which is even,
There are two middle terms, the \(\frac{2n}{2}\)^th i.e. T_n and T_n+1
T_n = ^2n-1C_n-1x^n-1 and T_n+1 = ^2n-1C_nxⁿ 
Coefficient of T_n = ^2n-1C_n-1 and coefficient of T_n+1 = ^2n-1C_n
Sum of the coefficients of the two middle terms in the expansion of
(1+x)^2n-1 = ^2n-1C_n + ^2n-1C_n-1 = ²ⁿC_n  --- (2)
RHS of(1) = RHS of (2), [∵ ⁿC_r+ⁿC_r-1 = ⁿ⁺¹C_r].

Let a and r be the first term and common ratio of the G.P.
\(\therefore\)  \({S}_{n}={a(1-r^n)\over1-r},n\epsilon N\)
LHS \(={{S}_{10}-{S}_{20}}^{2}=\left[ {a(1-{r}^{10})\over1-r}-{a(1-{r}^{20})\over1-r} \right]^{2}\)
\(={{a}^{2}\over{{(1-r)}^{2}}}[1-{r}^{10}-1+{r}^{20}]^2\)
\(={{a}^{2}\over{(1-r)}^{2}}.{r}^{20}{({r}^{10}-1)}^{2}={{a^2.{r}^{20}.{({r}^{10}-1)}^{2}}\over{{(1-r)}^{2}}}\)
RHS = S₁₀ (S₃₀ - S₂₀)
\(={a(1-{r}^{10}\over1-r)}\left[ {a(1-{r}^{30})\over1-r}-{{a(1-{r}^{30})}\over{1-r}} \right]\)
\(={{a^2}\over{(1-r^2)}}(1-{r}^{10})[1-{r}^{30}-1+{r}^{20}]={{a^2(1-{r}^{10})}\over{{(1-r)}^{2}}}.{r}^{20}(1-{r}^{10})\)
\(={{a^3.{r}^{20}(1-{r}^{10})^2}\over{{(1-r)}^{2}}}={{{a}^{2}.{r}^{20}{({r}^{10}-1)}^{2}}\over{{(1-r)}^{2}}}\)
\(\therefore\) LHS = RHS.

Given T_p+1 = 2.T_q+1
\(\Rightarrow\)   a+(+1-1)d = 2[a+(q+1-1)d]    [ \(\because\) T_n = a + ( n - 1) d ]
\(\Rightarrow\)  a + pd = 2a +2qd
\(\Rightarrow\) a = ( p - 2q ) d   ...(1)
 Now T_3p+1 = a+ ( 3p + 1 - 1 )d = a + 3pd
= ( p - 2q ) d + 3pd   (using (1))
= 4pd - 2qd
= 2d (2p-q) ...(2)
Also T_p+q+1 = a + (p+q+1-1)d
= a+(p+q)d
= (p-2q)d + (p+q)d   (using (1))
= d(p-2q+p+q) = d(2p-q) ...(3)
From (2) and (3), T_3p+1 = 2. T_p+q+1.

General term in (1+x)ⁿ is
T_r+1 = nC_r(1)^n-r.x^r = nC_rx^r
\(\therefore\) Co-efficient of T_r+1 is n C_r.
Let the Co-efficients of T_r+1, T_r+2 and T_r+3 be in the ratio 1 : 3 : 5
\(\therefore\) nC_r : nC_r+1 : nC_r+2 = 1: 3: 5
\(\Rightarrow\)  \({{n{C}_{n1}}\over{n{C}_{r}}}={3\over 1}\)   ....(1) 
and \({n{C}_{n+2}\over{C}_{n+1}}={5\over3}\)    ...(2)
From (1)  \(\rightarrow{\frac { \frac { n! }{ (r+1)!(n-r-1)! } }{ \frac { n! }{ r!(n-r)! } } }=3\)
\(\Rightarrow\)   \(\frac{r!(n-r)!}{(r+1)!(n-r-1)!}=3\)
\(\Rightarrow\)      
\(\Rightarrow{n-r\over+1}=3\)
Similarly (1) implies
\(\frac { \frac { n! }{ (r+2)!(n-r-2)! } }{ \frac { n! }{(r+1)!(n-r-1)! } } =\frac{5}{3}\)
\(\frac{(r+1)!(n-r-1)!}{(r+2)!(n-r-2)!}=\frac{5}{3}\Rightarrow\)  
\(\Rightarrow{n-r-1 \over r+2}={5\over 3}\Rightarrow 3n-3t-3=5r+10\)
\(\Rightarrow\)  3n - 8r - 13 = 0
(3) X 2 \(\rightarrow\) 2n-8r-6 = 0