\((2+2 x)^{10} \text { Term containing } x^{6} \text { is }\)
\({ }^{10} \mathrm{C}_{6}(2)^{10-6}(2 x)^{6}={ }^{10} \mathrm{C}_{6} 2^{4} 2^{6} x^{6}\)
\(\text { Coefficient of } x^{6} \text { is }{ }^{10} \mathrm{C}_{6} 2^{10}\)

\(\mathrm{a} =1, \quad \mathrm{~d}=6, \quad \mathrm{r}=\frac{1}{2} \)
\(\mathrm{~S}_{\infty} =\frac{a}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1+\mathrm{r})^{2}} \)
\(=\frac{1}{1-\frac{1}{2}}+\frac{6 \times \frac{1}{2}}{\left(\frac{1}{2}\right)^{2}} \)
\(=2+(3 \times 4)=14 \)

\(\frac{1}{\sqrt{1}+\sqrt{3}} =\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{2} \)
\(\frac{1}{\sqrt{3}+\sqrt{5}} =\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
\(=\frac{\sqrt{5}-\sqrt{3}}{2} \)
\(\text { Sum to } \mathrm{n} \text { terms }=\frac{(\sqrt{3}-1)}{2}+\frac{(\sqrt{5}-\sqrt{3})}{2}+\ldots . .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}\right)\)
\(=\frac{\sqrt{2 n+1}-1}{2}\)

\(a+b= 16, a b=16, x=\frac{2 a b}{a+b} \)
\(x=\frac{2 \times 16}{16}\)
\(x=2\)

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 

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.

Suppose the loan in cleared in n months. Clearly the amount forms an. A.P. with a = 20 and d = 15
∴ Sum of the amounts = 3250
S_n = 3250

\(⇒\ {n\over2}[2a + (n -1)d]=3250\)
\(⇒\ {n\over2}[40+(n-1)15]=3250\)
⇒ n(40 + 15n - 15) = 6500
⇒ n (15n + 25) 6500
⇒ 15n² + 25n = 6500
⇒ 15n² + 25n = 6500
⇒ 3n² + 5n - 1300 = 0
⇒ (n - 20) (3n + 65) = 0
⇒ n = 20 or \(n={-65\over 3}\) which is not possible
∴ n = 20
Thus, the amount is cleared in 20 months.

Given equation is (q - r).x²+ (r - p)x +p - q = 0
a = q - r, b = r - p, c = p - q
Since the roots of the quadratic equation are equal, b² - 4ac = 0
\(\Rightarrow\) (r - p)² - 4(q - r)(p - q) = 0
\(\Rightarrow\) r² +p² - 2rp - 4(pq - q² - rp + rq) = 0
\(\Rightarrow\) r² + p² - 2rp - 4pq + 4q² + 4rp - 4rq = 0
\(\Rightarrow\) r² + p² + 4q² + 2rp - 4pq - 4rq = 0
\(\Rightarrow\) (r + p - 2q)² = 0
\(\Rightarrow\) r + p - 2q = 0
\(\Rightarrow\) 2q = r + p
\(\Rightarrow\) q - p = r - p
\(\Rightarrow\) common difference is equal for p, q, r
Henc p, q, r are in A.P.

In (1 + x)ⁿ, (r + 1)^th term from the beginning.
T_r+1 = nC_r 1^n-r. x^r = nC_rx^r   ....(1)
Its co-efficient is nC_r
In (1 + x)ⁿ, there are (n + 1)terms
So, the (r +1)^th term from the end will have (n + 1) - (r + 1) = n - r terms
∴ T_n-r+1 = nC_n-r 1^n-(n-r).x^n-r = nC_n-rx^n-r  ...(2)
Its Co-efficient is  nC_n-r
From (1) and (2), the Co-efficient of (r + 1)th term from the beginning and from the end are equal

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 \(a=\frac { 12 }{ n } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)

In (1+x)ⁿ, n = n, x = 1, A = x
general terms t_r+1 = nC_r, X^n-r a^r
t_r+1 = nC_r, (1)^n-r a^r
t_r+1 = nC_r x^r
∴ Co-efficient of X^r is nC_r
Putting r = n -r in (1) we get
T_n-r+1 = nC_n-r X^n-r
Co-efficient of x^n-r is nC_n-r
But nC_r = nC_n-r Using the property of combination
ஃ Coefficients of x^r and Co-efficients of x^n-r  are equal

Given (x + y)ⁿ
If n is odd, the two middle terms in (x +y)ⁿ are  \({T_{n-1}\over 2}\ and \ {T_{n+1}\over 2}\)
\({T_{n-1}\over 2}=nC_{n+1\over 2}x^{n+1\over 2}y^{n-1\over 2}\ and\ {T_{n+1}\over2}=nC_{n-1\over2}x^{n-1\over 2}y^{n+1\over 2}\)
The co-efficients of middle terms are \(nC_{n+1\over2}\ and \ nC_{n-1\over2}\)
\(nC_{n+1\over2}=nC_{n-1\over2}⇒{n+1\over2}={n-1\over 2}\ or\ {n+1\over2}+{n-1\over 2}=n\)
[∴ nC_x = nC_y ⇒ x = y or x +y = n]
\(⇒\ {n+1\over 2}={n-1\over 2}0=2\) which is not possible
Also, \({n+1\over 2}+{n-1\over 2}=n⇒{n+n+1-1\over 2}=n\)
⇒ \({2n\over 2}=n⇒ n=n\)
∴ \(nC_{n+1\over 2}=nC_{n-1\over 2}\). Hence the coefficients of two middle terms are equal

Let T_n be the nth term of the given sequence.
Given sequence is \({1\over1},{4\over3},{7\over9},{10\over27}....\)
Consider the terms in the numerator
1, 4, 7, 10,...
Here a = 1, d = 3
The terms in the denominator are  \({1\over3^0},{1\over3^1},{1\over 3^2}\), which is a G.P with  \(r={1\over3}\)
∴ The given sequence can be written in the form of a, (a + d)r, (a + 2d)r²,(a + 3d), r³, ...
This is an arithmetic - geometric progression.
∴ T_n = [a+(n-1)d]r^n-1
\(=[1+(n -1)3]\left(1\over3\right)^{n-1}\)
\(=({1+3n-3})\left(1\over 3^{n-1}\right)={3n-2\over 3^{n-1}}\)
\(∴\ T_n={3n-2\over 3^{n-1}}\)
Let S_n be the sum to n terms of the given sequence
\(S_n=\sum_{k=1}^n{3k-2\over 3^{k-1}}\)
\(={\sum_{k=1}^n3k-2.{1\over{\sum_{k=1}^n}3^{k-1}}}\)
\(= 3[1+ 2 + 3+ ...+ n] - 2n \left[ 1\over3^0+3^2+...+3^{r-1}\right]\)
\(=\left[ 3{n(n+1)\over2}-2n\right]\left[ 1\over 1\left(3^n-1\over 3-1\right)\right]\)
 \(=\left[{3n^2+3n\over2}-2n\right]\left[2\over 3^n-1\right]={3n^2+3n-4n\over2}\times{2\over3^n-1}\)
\(\frac { { 3n }^{ 2 }-n }{ { 3 }^{ n }-1 } =\frac { n\left( n-1 \right) }{ { 3 }^{ n }-1 } \)

=[(x-a)ⁿ = xⁿ + ⁿC₁x^n-1(-a)¹+ⁿC₂x^n-1(-a)²+.....(-a)ⁿ]
= \(​​​​​​​​\left({ 2x }^{ 2 } \right) ^{ 3 }+3C_{ 1 }\left( { 2x }^{ 2 } \right) ^{ 2 }\left( \frac { 3 }{ x } \right) ^{ 1 }+{ 3C }_{ 2 }\left( { 2x }^{ 2 } \right) ^{ 1 }\left( \frac { 3 }{ x } \right) ^{ 2 }+\left( -\frac { 3 }{ x } \right) ^{ 3 }\)
= \({ 8x }^{ 6 }+3\left( { 4x }^{ 4 } \right) \left( -\frac { 3 }{ x } \right) +\frac { 3\times 2 }{ 2\times 1 } \left( { 2x }^{ 2 } \right) \left( \frac { 9 }{ { x }^{ 2 } } \right) -\frac { 27 }{ { x }^{ 3 } } \)
= \({ 8x }^{ 6 }-{ 36x }^{ 3 }+54-\frac { 27 }{ { x }^{ 3 } } \)

\(\sqrt [ 3 ]{ 126 } ={ (125) }^{ 1/3 }=(125+1)^{ 1/3 }=\left\{ 125\left( 1+\frac { 1 }{ 125 } \right) \right\} ^{ 1/3 }=(125)^{ 1/3 }\left[ 1+\frac { 1 }{ 125 } \right] ^{ 1/3 }\)
\(=5\left[ 1+\frac { 1 }{ 3 } \times \frac { 1 }{ 125 } +... \right] \left( \therefore \frac { 1 }{ 125 } <1 \right) =5\left[ 1+\frac { 1 }{ 3 } (0.008) \right] =5(1+0.002666)=5.01\)

As n = 7 which is odd, the terms containing x⁴y³ and x³y⁴ are the two middle terns.
They are ⁷C₃ x⁴y³ and ⁷C₄x³y⁴ which are equal 35x⁴y³ and 35x³y⁴.

Let T_n be the n^th term of the sequence
Then, given T₁ = 3, T₂ = 3, T₃ = 6 and 
T_n = T_n-1 + T_n-2, n > 2.
T₃ = T₂ + T₁ = 3 + 3 = 6
T₄ = T₃ + T₂ = 6 + 3 = 9
T₅ = T₄ + T₃ = 9 + 6 = 15.