Given \(\sqrt [ 3 ]{ 1001 } ={ \left( 1000+1 \right) }^{ \frac { 1 }{ 3 } }={ \left( 1000 \right) }^{ \frac { 1 }{ 3 } }{ \left( 1+\frac { 1 }{ 1000 } \right) }^{ \frac { 1 }{ 3 } }\)
\(={ 10 }^{ 3\times \frac { 1 }{ 3 } }{ \left[ 1+\frac { 1 }{ 1000 } \right] }^{ \frac { 1 }{ 3 } }\)
\(\sqrt [ 3 ]{ 1001 } =10{ \left( 1+.001 \right) }^{ \frac { 1 }{ 3 } }\)
\(=10\left[ 1+\frac { .001 }{ 3 } +\left( \frac { 1 }{ 3 } \right) \left( -\frac { 2 }{ 3 } \right) \left( \frac { .000001 }{ 2 } \right) \right] app\)
\(=10\left[ 1+.00033-\frac { .000001 }{ 9 } \right] app\)
= 10 [1.00033 - .00000011] app
= 10 [1.000329]
= 10 [1.00033]
\(\\ \\ { \left( 1000 \right) }^{ \frac { 1 }{ 3 } }\cong 10.0033\)

LHS = \({ \left( { x }^{ 3 }+6 \right) }^{ \frac { 1 }{ 3 } }-{ \left( { x }^{ 3 }+3 \right) }^{ \frac { 1 }{ 3 } }\)
\(={ x }^{ 3\times \frac { 1 }{ 3 } }{ \left( 1+\frac { 6 }{ { x }^{ 3 } } \right) }^{ \frac { 1 }{ 3 } }-{ x }^{ 3\times \frac { 1 }{ 3 } }{ \left( 1+\frac { 3 }{ { x }^{ 3 } } \right) }^{ \frac { 1 }{ 3 } }\)
\(=x\left[ 1+\frac { 1 }{ 3 } \left( \frac { 6 }{ { x }^{ 3 } } \right) \right] -x\left[ 1+\frac { 1 }{ 3 } \left( \frac { 3 }{ { x }^{ 3 } } \right) \right] \)
\(=x+\frac { 2 }{ { x }^{ 2 } } -x-\frac { 1 }{ { x }^{ 2 } } \)
\(=\frac { 2 }{ { x }^{ 2 } } -\frac { 1 }{ { x }^{ 2 } } =\frac { 1 }{ { x }^{ 2 } } =RHS\)
Hence proved.

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.
Using the given in information, we have \(\frac{S_m}{S_n}=\frac{m^{2}}{n^{2}}\)
∴ \(\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m{2}}{n^{2}}\Rightarrow \frac{2a+(m-1)d}{2a+(n-1)d} \overset{-}{n}\)
⇒ 2an + (mn - n) d = 2am + (mn - m) d
⇒ 2an - 2am = (mn - m - mn + n) d
⇒ 2a (n - m) = (n - m) d⇒ d = 2a, [n-m≠0, as n≠m]
Now, \(\frac{a_{m}}{a_{n}}=\frac{a+(m-1d)}{a+(n-1)d}=\frac{a+(m-1).2a}{a+(n-1).2a}=\frac{a(1+2m-2)}{1+2n-2}=\frac{2m-1}{2n-1}\)
Hence, the required ratio is (2m - 1) : (2n - 1) 

Given that, Sum of the coefficients in the expansion of  (x+y)ⁿ = 4096
 ∴ ⁿC₀ + ⁿC₁ + ⁿC₂ +..+ⁿC_n = 4096
[∴ Sum of binomial coefficients in the expansion of (x + a)ⁿ is 2ⁿ]
⇒ 2ⁿ = 4096 = 2¹²
⇒ n = 12 (even)
So the greatest coefficient = Coefficient of the middle term \((\frac{n}{2}+1)\)th term 
= Coefficient of the middle term \((\frac{12}{2}+1)\)th term
= Coefficient of the 7^th term
Hence the greatest coefficient = ¹²C₆=\(\frac{(12)!}{6!(12-6)!}=\frac{(12)!}{6!6!}=\frac{12\times11\times10\times9\times8\times7}{6\times5\times4\times3\times2\times1}=924\) 

Let T_r+1 be the term in whichx³² and x^-17 occurs,
\(\therefore T_{r+1}= {^{15}C_{r}}.(x^{4})^{15-r}(-\frac{1}{x^{3}})^{r}\)
= \({^{15}C_{r}},(-1)^{r},x^{60-4r},x^{-3r}={^{15}C_r},(-1)^{r}, x^{60-7r}\)
(i) Since x³² occurs in this term
∴ Exponent of x = 32
⇒ 60 - 7r = 32 ⇒ 7r = 28
∴ r = 28 ÷ 7 = 4
∴ Coefficient ofthe term containing x³² is = ¹⁵C₄(-1)⁴ = 1365
(ii) Since x^-17occurs in this term
∴ Exponent of x = -17
⇒ 60-7r = -17
⇒ 7r = 77,  ∴ r = 11
∴ Coeffiicciient of the term containing x^-17=¹⁵C₁₁(-1)¹¹= -¹⁵C₁₁(-1)¹¹ = -¹⁵C_15-11 = -¹⁵C₄= -1365.

Here a = 1 d = 3 and r \(={1\over5}\)
\(s_\infty={a\over 1-r}+{dr\over (1-r)^2}\)
\(={1\over 1-{1\over5}}+{3\times{1\over 5}\over({1-{1\over 5}})^2}\)
\(={5\over 4}+({3\over 5})({25\over 16})={35\over 16}\)

Let the H.P. be \(\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},...\)
\(\therefore { T }_{ m }=\frac { 1 }{ a+\left( m-1 \right) d } =n\)   and \(\therefore { T }_{ n }=\frac { 1 }{ a+\left( n-1 \right) d } =m\)
a + (m - 1)d = \(\frac{1}{n}(1)\) and a + (n - 1)d = \(\frac{1}{m}\)
(1) - (2) \(\Rightarrow\) (m - 1 - n + 1)d = \(\frac{1}{n}-\frac{1}{m}\) \(\Rightarrow\) (m - n) d = \(\frac{m-n}{mn}\Rightarrow d=\frac{1}{mn}\)
\({ T }_{ p }=\frac { 1 }{ a+\left( p-1 \right) d } =\frac { 1 }{ \frac { 1 }{ mn } +\left( b-1 \right) \frac { 1 }{ mn } } =\frac { mn }{ 1+p-1 } \)
\({ T }_{ p }=\frac { mn }{ p } \)

Let T₁,T₂,T_3... be a sequence which is A.P as well as G.P
Let Tn = a+(n-1)d for all n ∈ N
∴ The sequence is a, a+d, a+2d,...
Now, this is also a G.P
\(\therefore \frac { { T }_{ n+1 } }{ { T }_{ n } } =\frac { { T }_{ n+2 } }{ { T }_{ n+1 } } \)    for all n ∈ N
⇒ \(\frac { a+nd }{ a+(n-1)d } =\frac { a(n+1)d }{ a+nd } \)
⇒ \((a+nd)^{ 2 }=(a+nd+d)(a+nd-d)\)
⇒ \((a+nd)^{ 2 }=(a+nd)^{ 2 }-{ d }^{ 2 }\)
⇒ d² = 0 ⇒ d = 0
∴  The sequence is a+0, a+2(0), a+3(0) +...
⇒ a,a,a...
∴ Only a constant sequence can be both A.P and G.P.

Let r be the common ratio of the G.P
Here a = 1 and T₃+T₅ = 90
⇒ ar²+ar⁴ = 90     [∵ T₃ = a.r²,T₄ = a.r³]
⇒ r²+ r⁴ = 90   (∵ a = 1)
⇒ r²+ r⁴-90 = 0
⇒ r² = \(\frac { -1\pm \sqrt { 1+360 } 2 }{ 2 } \)   \(\left[ \because \ x=\frac { x={ -b\pm \sqrt { b^{ 2 }-4ac } } }{ 2a } a=1,b=1,c=-90 \right] \)
⇒ r² = \(\frac { -1\pm 19 }{ 2 } \)
⇒ r² = \(\frac { -1+19 }{ 2 } \)  or \(\frac { -1-19 }{ 2 } \)
⇒ r² = 9 ⇒ r = -3,     [∵ r² = -10 is impossible ]
∴ r = 3 or -3