It is given that,
f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)
f (1) = 3
Taking x = y = 1 in (1), we obtain
f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9
Similarly,
f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27
f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81
∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.
\(\text{It is known that, }S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}\)
\(\text {It is given that, } \sum_{x=1}^{n} f(x)=120 \therefore 120=\frac{3\left(3^{n}-1\right)}{3-1}\)
\(\Rightarrow 120=\frac{3}{2}\left(3^{n}-1\right)\)
\(\Rightarrow 3^{n}-1=80\)
\(\Rightarrow 3^{n}=81=3^{4}\)
\(\therefore n = 4\) 
Thus,the value of n is 4.

Let \(\frac {a}{r}\) a, ar be the first three terms of the G.P. Then
\(\frac{a}{r}+a r+a=\frac{13}{12} \)
and  \(\left(\frac{a}{r}\right)(a)(a r)=-1\)
From (2), we get a³ = – 1, i.e., a = – 1 (considering only real roots)
Substituting a = –1 in (1), we have
\(-\frac{1}{r}-1-r=\frac{13}{12} \text { or } 12 r^{2}+25 r+12=0\)
This is a quadratic in r, solving, we get \(r=-\frac{3}{4} \text { or }-\frac{4}{3}\)
Thus, the three terms of G.P. are : \(\frac{4}{3},-1, \frac{3}{4} \text { for } r=\frac{-3}{4} \text { and } \frac{3}{4},-1, \frac{4}{3} \text { for } r=\frac{-4}{3}\)

Here a = 1 and r = \(\frac{2}{3}\) therefore
\(\mathrm{S}_{n}=\frac{a\left(1-r^{n}\right)}{1-r}=\frac{\left[1-\left(\frac{2}{3}\right)^{n}\right]}{1-\frac{2}{3}}=3\left[1-\left(\frac{2}{3}\right)^{n}\right] \)
In particular,\(\mathrm{S}_{5}=3\left[1-\left(\frac{2}{3}\right)^{5}\right]=3 \times \frac{211}{243}=\frac{211}{81} .\)

Total cost of the tractor = Rs12000
Cash amount paid = Rs 6000
Balance amount = Rs12000 - Rs 6000 = Rs 6000
Interest of 1st instalment =Rs\(\frac { 6000\times 12\times 1 }{ 100 } \)=Rs 720
Amount of 1st instalment =Rs500+Rs720=Rs1220
Interest of second instalment =Rs \(\frac { 5500\times 12\times 1 }{ 100 } \)=Rs 660
Amount of second instalment =Rs 500+Rs 600=Rs1160
Interest of the 3rd instalment \(=\frac { 5000\times 12\times 1 }{ 100 } \)=Rs 600
Amount of third instalment = 500 + 600 = Rs1100
The sequence of instalment is 1220, 1160, 1100 ...
Here a = 1220 d = 1160 - 1220 = - 60 and n = 12
∴ S_n=\(\frac { n }{ 2 } \)[2a+(n-1)d]
∴ S₁₂=\(\frac { 12 }{ 2 } \)[2 x 1220+(12-1) x -60]
=6[2400-660]=Rs 10680
Thus total cost of tractor =10680+6000=Rs 16680

Given,T_p = a \(\Rightarrow\) AR^p-1  = a .....(i)
T_q = b \(\Rightarrow\)  AR^q-1  = b.....(ii)
and  T_r = c \(\Rightarrow\)  AR^r-1 = c ....(ii)
LHS = a^q-r b ^r-p b ^p-q 

Let the GP be a,ar,ar₂,ar₃,....
Given a = 1 and T₃ + T₅ = 90
\(\therefore\) ar2 + ar⁴ = 90 \( \Rightarrow\) r² + r⁴ = 90
r⁴ + r2 - 90 = 0   [as a =1] 
( r² + 10 ) ( r² - 9 ) = 0 \( \Rightarrow\) r = \(\pm\) 3
[ \(\because\) r² = -10 is a complex number]

\(\therefore\) A.M. of a and b = \(\frac{a+b}{2}\)
G.M of a and b = \(\sqrt{ab}\)
\(\therefore\)  \(\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}\)
By componendo and dividendo, we get
\(\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{m+n}{m-n}\)
\(\Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{m+n}{m-n}\)
\(\Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}}\)
Again by componendo and dividendo, we have:
\(\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\)
\(\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\)
Squaring both sides,
\(\frac{a}{b}=\frac{(\sqrt{m+n}+\sqrt{m-n})^2}{(\sqrt{m+n}-\sqrt{m-n})^2}\)
\(\Rightarrow \frac{a}{b}=\frac{m+n+m-n+2\sqrt{(m+n)+\sqrt{m-n}}}{m+n+m-n-2\sqrt{(m+n)(m-n)}}\)
\(\Rightarrow \frac{2m+2\sqrt{m^2-n^2}}{2m-2\sqrt{m^2-n^2}}\)
\(\Rightarrow \frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}\)
Thus, a: b = \((m+\sqrt{m^2-n^2}):(m-\sqrt{m^2-n^2})\)

Let a, ar, ar² be three numbers in G.P
\(\therefore\) a +ar +ar² = 56
\(\Rightarrow\) a (1 + r + r²) = 56 ...(1)
According to the given condition if we subtract 1,7,21 from a, ar, ar² respectively then resulting numbers are also in AP.
\(\because\)  a - 1, ar - 7, ar² - 21 are in AP.
\(\therefore\)  (ar - 7) - (a - 1) = (ar² - 21) - (ar - 7)
\(\Rightarrow\) ar - 7 - a + 1 = ar² - 21 - ar + 7
\(\Rightarrow\) ar - a - 6 = ar² - ar -14
\(\Rightarrow\) ar² - 2ar + a = 8
\(\Rightarrow\) a (r² - 2r + 1) = 8
Dividing (1) by (2) we get
\(\frac { a\left( 1+r+{ r }^{ 2 } \right) }{ a\left( { r }^{ 2 }-2r+1 \right) } =\frac { 56 }{ 8 } \)
\(\Rightarrow\) r² + r + 1 = 7r² - 14r + 7
\(\Rightarrow\) 6r² - 15r + 6 = 0
\(\Rightarrow\) 2r² -5r + 2 = 0
\(\Rightarrow\) r = \(\frac { -(-5)\pm \sqrt { \left( 5 \right) ^{ 2 }-4\times 2\times 2 } }{ 2\times 2 } \)
\(\Rightarrow\) r = \(\frac { 5\pm \sqrt { 25-16 } }{ 4 } \) \(\Rightarrow\) r = \(\frac { 5+3 }{ 4 } \)
Either r = \(\frac { 5+3 }{ 4 } \) i.e r = \(\frac { 8 }{ 4 } \) \(\Rightarrow\)r = 2
or r = \(\frac { 5-3 }{ 4 } \) i.e r = \(\frac { 2 }{ 4 } \) \(\Rightarrow\) r = \(\frac { 1 }{ 2 } \)
Now, putting r = 2 in (1), we get
a (1 + 2 + 2²)= 56  a = 8
:. The required numbers are: 8, 16, 32
Putting r = \(\frac { 1 }{ 2 } \) in (1) we get
\(a\left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) =56\)
\(\Rightarrow\) \(a\left( \frac { 4+2+1 }{ 4 } \right) =56\)
\(\Rightarrow\) \(a\left( \frac { 7 }{ 4 } \right) =56\)
\(\Rightarrow\) a = 56 x \(\frac { 4 }{ 7 } =32\)
\(\therefore\) The requireds numbers are 32,16,8

Given, \(\frac { { a }^{ n+1 }+{ b }^{ n+1 } }{ { a }^{ n }+{ b }^{ n } } =\frac { { a }^{ \frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } } }{ 1 } \)
\(\Rightarrow { a }^{ n+1\quad }+{ b }^{ n+1 }={ a }^{ n+\frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } }+{ a }^{ \frac { 1 }{ 2 } }{ b }^{ n+\frac { 1 }{ 2 } }\)
\(\Rightarrow ({ a }^{ n+1 }-{ a }^{ n+\frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } })+({ b }^{ n+1 }-{ a }^{ \frac { 1 }{ 2 } }{ b }^{ n+\frac { 1 }{ 2 } })=0\)
\(\Rightarrow ({ a }^{ n+\frac { 1 }{ 2 } }-{ b }^{ n+\frac { 1 }{ 2 } })({ a }^{ \frac { 1 }{ 2 } }-{ b }^{ \frac { 1 }{ 2 } })=0\)
\(\Rightarrow { a }^{ n+\frac { 1 }{ 2 } }- { b }^{ n+\frac { 1 }{ 2 } }=0\)

Let 'a' be the first term and 'r' be the common ratio of the given A.P.
Here, a₅ = p \(\Rightarrow \)   ar⁴ = p....(1)
 a₈ = q  \(\Rightarrow \)   ar⁷ = q..(2) 
a₁₁ = s \(\Rightarrow \)  ar¹¹ = s ....(3) 
Squaring both sides of equation (2), we get
q² = (ar⁷)² \(\Rightarrow \) q²= a²r¹⁴
\(\Rightarrow \)q²=(ar⁴)(ar¹⁰)
\(\Rightarrow \)q²=ps [\(\therefore \) p=ar⁴ and s = ar¹⁰]

Let a and b be roots of required quadratic equation.
Then A.M = \(\frac { a+b }{ 2 } =8\) and G.M = \(\sqrt { ab } \) = 5
\(\therefore\) a + b = 16 and ab = 25
If S is sum of roots and P is product of roots then quadratic equation is x² - Sx + P = 0
\(\therefore\) Required quadratic equation is = x² - 16x + 25 = 0

Here, \({ a }_{ n }=\frac { n(n-2) }{ n+3 } \)
Putting n = 20, we have
\({ a }_{ 20 }=\frac { 20(20-2) }{ 20+3 } =\frac { 20\times 18 }{ 23 } =\frac { 360 }{ 23 } \)
Thus, the 20^th term of sequence is \(\frac { 360 }{ 23 } \)

Given T₄= x \(\Rightarrow \) ar^4-1=x \(\Rightarrow \) ar³=x ...(i)
T₁₀=y\(\Rightarrow \) ar^10-1=y\(\Rightarrow \) ar⁹=y..(ii)
T₁₆=z\(\Rightarrow \) ar^16-1=y\(\Rightarrow \) ar¹⁵=z .(iii)
Now , on multiplying Wq(i) by Eq(iii) , we get
\({ ar }^{ 3 }\times { ar }^{ 15 }=x\times z\Rightarrow { a }^{ 2 }{ r }^{ 3+15 }=xz\)
\(\Rightarrow { a }^{ 2 }{ r }^{ 18 } \Rightarrow ({ ar }^{ 9 })^{ 2 }=xz\)
\(\Rightarrow { y }^{ 2 }=xz\)
There,x,y,z are in G.p.

Given  a,b,c,d arein GP.
\(\therefore\) \(\frac{b}{a}\) = \( \frac{c}{b}\) = \(\frac{d}{c}\) = r (say)
\(\Rightarrow\) b  = ar, c = br, d = cr
\(\Rightarrow\) b = ar, c = ( ar ) r, d =( br ) r
\(\Rightarrow\) b = ar,  c=ar², d = br²
\(\Rightarrow\) b = ar, c =ar², d ( ar ) r²= ar³ ....(i)
Now,we have to prove that 
( a² + b² + c² )( b² + c² + d² ) = ( ab + bc + cd )²
LHS =  ( a² + b² + c² )( b² + c² + d² )
= ( a² + a²r² + a²r⁴ )( a²r² + a²r⁴ + a²r⁶ )
= a²( 1 + r² + r⁴ )a²r²( 1+ r² + r⁴ )
= a⁴r² ( 1 + r² + r⁴ )²
= [ a²r ( 1 + r² + r⁴ ) ]²
= [ a²r + a²r³ + a²r⁵ ]²
= [ a.ar + ar.ar² + ar².ar³ ]²
= [ ab+bc+cd ]²  [ from Eq.( i ) ]
=RHS

Let a be the first term and r be the common ratio of the given GP.
According to the question,
T₄ = (T₂)² and a = -3
\(\because \) T₄ = (T₂)²  \(\therefore \)  ar³ = (ar)² 
\(\Rightarrow\)  -3 r³ = (-3)² r²  [\(\because \) a = -3]
\(\Rightarrow\)  r = -3
Now, T₇ = ar⁶ = -3(-3)⁶ = -3 x 729 = -2187