Here a = 1, r = 4 and S_n = 5461
We know that S_n=\(\frac { a\left( { r }^{ n }-1 \right) }{ r-1 } \)
5461=\(\frac { 1.\left( { 4 }^{ n }-1 \right) }{ 4-1 } \)
=4ⁿ-1 =16383 \(\Rightarrow \)4ⁿ=16384 \(\Rightarrow \) 4ⁿ=4⁷\(\Rightarrow \)n=7.

Let a be the first term and r be the common ratio of given G.P. Then
a₇=8a₄ and a₅=48
\(\therefore \) ar⁶=8ar³and ar⁴=48
\(\Rightarrow \)r³=8\(\Rightarrow \) r=2
Putting r = 2 in are⁴=48
\(\therefore \)  a(2)⁴=48 \(\Rightarrow \) a=\(\frac { 48 }{ 16 } \) \(\Rightarrow \) a=3.

Let a be the first term and r be the common ratio of the given GP.
According to the question,
T₅ = 48  \(\Longrightarrow \)  ar⁴ = 48.......(i)
and T₈ = 384  \(\Longrightarrow \)  ar⁷ = 384......(ii)
On dividing Eq.(ii) by Eq. (i), we get
\(\Longrightarrow \)  \(\frac { a{ r }^{ 7 } }{ a{ r }^{ 4 } } \) = \(\frac { 384 }{ 48 } \)  \(\Longrightarrow \)  r³ = 8 \(\Longrightarrow \)  r = 2
On putting r = 2 in Eqⁿ (i), we get a = 3
\(\therefore \)   3, 6, 12, .......are in GP.

Here a=2,\(r=\frac { 4 }{ 2 } =2\) and n=10
\(\because \ { S }_{ 10 }=\frac { 2({ 2 }^{ 10 }-1) }{ 2-1 } =2({ 2 }^{ 10 }-1)=2(1024-1)=2046\)

Let the two numbers be a and b such that a > b.
Then, \(\frac { a+b }{ 2 } \) = 34 and \(\sqrt { ab } \) = 16
Now, (a - b)² = (a + b)² -4ab \(\Rightarrow \) a - b =60
64 and 4

Given series is  4+44+444+..n terms.
On taking comment 4, we get
S_n=4(1+11+111+...n terms)
On multiplying and during each tern of RHS by 9, we get
S_n=\(\frac { 4 }{ 9 } (9+99+999+...)\) 
\(\Rightarrow { S }_{ n }=\frac { 4 }{ 9 } [(10-1)+(100-1)+(1000-1)+\)...n terms
\(\Rightarrow { S }_{ n }=\frac { 4 }{ 9 } [10+100+1000+...\)n terms)-(1+1+1+...n terms]
For first series,10+100+1000+..n trms
\(r=\frac { 100 }{ 10 } =10>1\)
S_n=10(1+10+10²+...n terms)
\(=10\left( \frac { 10^{ n }-1 }{ 10-1 } \right) \)
=\(\frac { 10 }{ 9 } ({ 10 }^{ n }-1)\)
Now from Eq.(i) we get
\({ S }_{ n }=\frac { 4 }{ 9 } \)\(\left[ \frac { 10 }{ 9 } ({ 10 }^{ n }-1)-n \right] =\frac { 40 }{ 81 } ({ 10 }^{ n }-1)-\frac { 4 }{ 9 } n\) 

Given,GP is \(3,\frac { 3 }{ 2 } ,\frac { 3 }{ 4 } \),...
Here,  a=3,\(r=\frac { 3 }{ 2 } \div 3=\frac { 1 }{ 2 } \)
Let n be the number of terms needed
Thus S_n=\(\frac { 3069 }{ 512 } \)\(​​\Rightarrow \frac { a(1-{ r }^{ n }) }{ 1-r } =\frac { 3069 }{ 512 }\quad [\because r<1]\)
\(\Rightarrow \frac { 3\left\{ 1-\frac { 1 }{ { 2 }^{ n } } \right\} }{ 1-\frac { 1 }{ 2 } } =\frac { 3069 }{ 512 } \)
\( \Rightarrow 6\left( 1-\frac { 1 }{ { 2 }^{ n } } \right) =\frac { 3069 }{ 512 } \Rightarrow 1-\frac { 1 }{ { 2 }^{ n } } =\frac { 3069 }{ 3072 }\)
\(\Rightarrow \frac { 1 }{ { 2 }^{ n } } =1-\frac { 3069 }{ 3072 } =\frac { 3072-3069 }{ 3072 } \\ \)
\(\Rightarrow \frac { 1 }{ { 2 }^{ n } } =\frac { 3 }{ 3072 } =\frac { 1 }{ 1024 } \Rightarrow 2^{ v }=1024\Rightarrow { 2 }^{ n }={ 2 }^{ 10 }\) 
On comparing the powers from both sides, we get 
n=10
hence,10 terms are needed to give the sum\(\frac { 3069 }{ 512 } \).

Let the two numbers be a and b such that a>b.
Given, a - b =12...(i)
and   AM - GM = 2
\(\Rightarrow \frac { a+b }{ 2 } -\sqrt { ab } =2[\because \quad AM=\frac { a+b }{ 2 } and\quad GM=\sqrt { ab } ]\)
\(\Rightarrow a+b-2\sqrt { ab } =4\)
\(\Rightarrow (\sqrt { a } -\sqrt { b } { ) }^{ 2 }=4\)
\(\Rightarrow \sqrt { a } -\sqrt { b } =\quad 2\)
Now, a - b = 12 ...(ii)
\(\Rightarrow (\sqrt { a } +\sqrt { b } )(\sqrt { a } -\sqrt { b } )=12\quad [\because \quad { x }^{ 2 }-{ y }^{ 2 }=(x-y)(x+y)]\)
\(\Rightarrow (\sqrt { a } +\sqrt { b } )\times (2)=12\)
On adding Eqs. (ii) and (iii), we get 
\(2\sqrt { a } =8 \Rightarrow \sqrt { a } = 4\)
\(\Rightarrow \) a = (4)² \(\Rightarrow \)  a = 16
Then, from Eq. (i) we get
16 - b = 12
\(\Rightarrow \)  b = 16 -12 = 4
Hence required numbers are 16 and 4.

\(6\times 0.1+6\times 0.11+6\times 0.111+....\)n terms
\(=\frac { 6 }{ 9 } [0.9+0.99+0.999+\)...n terms]
\(=\frac { 2 }{ 3 } \left[ \frac { 9 }{ 10 } +\frac { 99 }{ 100 } +\frac { 999 }{ 1000 } +...\quad n\quad terms \right] \)
\(=\frac { 2 }{ 3 } \left[ \left( 1-\frac { 1 }{ 10 } \right) +\left( 1-\frac { 1 }{ 100 } \right) +\left( 1-\frac { 1 }{ 1000 } \right) +...n\quad terms \right] \)
\(=\frac { 2 }{ 3 } \left[ (1+1+1+...n\quad terms)-(\frac { 1 }{ 10 } +\frac { 1 }{ { 10 }^{ 2 } } +\frac { 1 }{ { 10 }^{ 3 } } +...n\quad terms) \right] \) 
Ans.\(\frac { 2 }{ 3 } n-\frac { 2 }{ 27 } (1-10^{ -n })\)

a + ar + ar² = 21 \(\Rightarrow \) a² + ( ar )² + ( ar² )² = 189
Ans. ( 3,6,12 ) or ( 12,6,3 )

\(\frac{b}{a}\)= \(\frac{c}{b} \)= \(\frac{d}{c}\) = r
\(\Rightarrow \) b = ar, c = br = ar²,d = cr = ar³
Now, a²-b²=a²r²=a²(1-r²)
b²-c²=a²r²-a²r⁴= a²r²(1-r²)
c²-d²=a²r⁴-a²r⁶= a²r⁴(1-r²)
Therefore, \( \frac{b^2-c^2}{a^2-b^2}\) =  \(\frac{c^2-d^2}{b^2-c^2}\)
= r²

Let 'a' be the 1^st term and 'd' be the common difference of the given AP.
\(\therefore\) S_n = \(\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\therefore\) S_n = \(\frac { 2n }{ 2 } \left[ 2a+\left( 2n-1 \right) d \right] \)
\(\therefore\)  S_n = \(\frac { 3n }{ 2 } \left[ 2a+(3n-1)d \right] \)
Now S₂ - S₁ = \(\frac { 2n }{ 2 } \left[ 2a+(2n-1)d \right] -\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)
= \(\frac { n }{ 2 } \left[ 4a+4nd-2d-2a-nd+d \right] \)
= \(\frac { n }{ 2 } \left[ 2a+3nd-d \right] \)
= \(\frac { n }{ 2 } [2a+(3n-1)d]\)
3(S₂-S₁) = \(\frac { 3n }{ 2 } [2a+(3a-1)d]\) = S₃
[\(\therefore\) S₃ = \([2a+(3a-1)d]\)
Thus, S₃ = 3 (S₂-S₁)