Let a be the first term and r the common ratio of the G.P.
∴ S= a + ar + ar²+ ...+ ar^{n - 1}
\(=\frac{a(1-r^{n})}{1-r}\)  --- (1)
p =\(a\times ar \times ar^{2}\times...\times ar^{n-1}=a^{n}r^{1+2+3}+..+(n-1)=a^{n}r^{n{(n-1})/2}\)
∴ \(P^{2}=a^{2n}r^{n(n-1)}\)   --- (2)
\(R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^{2}}+....+\frac{1}{ar^{n-1}}\)
⇒ \(R=\frac{1}{a}.\frac{(1-\frac{1}{r^{n}})}{(1-\frac{1}{r})}=\frac{(r^{n}-1)}{(r-1)}.\frac{1}{ar^{n-1}}\)   [∵Here, r<1]
∴ \(\frac{S}{R}=a\frac{(1-r^{n})}{1-r}.\frac{r-1}{r^{n}-1} ar^{n-1}= a^{2}r ^{n-1}\)
∴ \((\frac{S}{R})^{n}=a^{2n}r^{n(n-1)}\)   --- (3)
From (2) and (3) we get \(P^{2}=(\frac{S}{R})^{n}\)

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 

\(\sqrt [ 3 ]{ x^3+7 } ={(x^3+7)}^{{1\over 3}}\)
\(={\left[ x^3\left( 1+{7\over x^3} \right) \right]}^{{1\over 3}}\)   (\(\left |{7\over x^3}\right |<1\) as x is large)
\(=x{\left( 1+{7\over x} \right)}^{1\over 3}\)
\(=x\left( 1+{1\over 3} \times {7\over x^3}+{{{1\over3}\left( {1\over 3}-1 \right)}\over{2!}} {\left( {{7\over x^3}} \right)}^{2} +......\right)\)
\(=x\left( 1+{7\over 3}\times{1\over x^3}-{49\over 9}\times{1\over x^6}+...... \right)\)
\(=x+{7\over 3}\times{1\over x^2}-{49\over 9}\times{1\over x^5}+...\)
\(\sqrt [ 3 ]{ x^3+4 } ={(x^3+4)}^{1\over 3}\)
\(={\left[ x^3\left( 1+{4\over x^3} \right) \right]}^{1\over 3}\)
\(=x\left( 1+{4\over x^3} \right)^{1\over 3}\)
\(=x{\left( 1+{1\over3}\times{4\over x^3}+{{1\over 3}\left( {1\over3}-1 \right)\over{2!}} {\left( {4\over x^3} \right)}^{2}+... \right)}^{1\over3}\)
\(=x+{4\over 3}\times{1\over x^3}-{16\over 9}\times{1\over x^5}+...\)
Since x is large, \({1 \over x}\) is very small and hence higher powers of \({1 \over x}\) are negligible.
Thus \(\sqrt [ 3 ]{ x^3+7 } =x+{7\over 3}\times{1\over x^2}\) and \(\sqrt [ 3 ]{ x^3+4 } =x+{4\over3}\times{1\over x^3}.\) Therefore
\(\sqrt [ 3 ]{ x^3+7 } -\sqrt [ 3 ]{x^3+4 } =\left(x+{7\over 3}\times{1\over x^2} \right)-\left( x+{4\over 3}\times{1\over x^2} \right)={1\over x^2}\)

(Clearly, we have to use the expansion of (1 +x)^-2. So, we have to write (3 + 2x) as \(3(1+{2x\over3})\) and proceed.)
\({1\over (3+2x)^2}={1\over 3^2(1+{2x\over3})^2}\)
\(={1\over 9}(1+{2x\over3})^{-2}\)
\(={1\over 9}(1+y)^{-2}\)    \((where \ y={2x\over 3})\)
\(={1\over 9}(1+2y+3y^3-4y^3+5y^4-....),\)   if |y|<1
\(={1\over 9}(1-2({2x\over3})+3({2x\over 3})^2-4({2x\over3})^3+5({2x\over3})^4-...),|{2x\over3}|<1\)
\(={1\over9}(1-{4\over3}x+{4\over3}x^2-{32\over27}x^3+{80\over81}x^4-......)\)
Thus,\({1\over (3+2x)^2}={1\over9}({4\over27}x+{4\over27}x^2-{32\over243}x^3+{80\over729}x^4-....,|x|<{3\over2})\)
The expansion is valid if |y| < 1. So, the expansion is valid if |x|<\(3\over2\)

Rs.270

\(\frac{x}{4}\)

\((x^2+\sqrt{1-x^2})^5\) = ⁵C₀(x²)⁵ \({(\sqrt{1-x^2})}^{0}\) + ⁵C₁(x²)⁴\({(\sqrt{1-x^2})}^{1}\)+⁵C₂(x²)³\({(\sqrt{1-x^2})}^{2}\)+⁵C₃(x²)²\({(\sqrt{1-x^2})}^{3}\)+⁵C₄(x²)¹\({(\sqrt{1-x^2})}^{4}\)+⁵C₅(x²)⁰\({(\sqrt{1-x^2})}^{5}\)
= x¹⁰+5x²\(\sqrt{1-x^2}\) +10x⁶(1-x²)+10x⁴(1-x²)\(\sqrt{1-x^2}+5x^2(1-x^2)^2+(1-x^2)^2(\sqrt{1-x^2})\)
\((x^2-\sqrt{1-x^2})^5\) =⁵C₀(x²)⁵\({(\sqrt{1-x^2})}^{0}\) - ⁵C₁(x²)⁴ \((\sqrt{1-x^2})^1\) +⁵C₂(x²)³\({(\sqrt{1-x^2})}^{2}\) -⁵C₃(x²)²\({(\sqrt{1-x^2})}^{3}\)+⁵C₄(x²)¹\({(\sqrt{1-x^2})}^{4}\) -⁵C₅(x²)⁰\({(\sqrt{1-x^2})}^{5}\)
\(={x}^{10}-5x^8\sqrt{1-x^2}+10x^6(1-x^2)-10x^4(1-x^2)\sqrt{1-x^2}+5x^2(1-x^2)^2-(1-x^2)^2(\sqrt{1-x^2})\)
Thus 
\((x^2+\sqrt{1-x^2})^5+(x^2+\sqrt{1-x^2})^5\) = 2[x¹⁰ + 10x⁶ (1 - x²) + 5x² (1-x²)²]
= 2[x¹⁰ + 10x⁶ -10x⁸+ 5x² (1- 2x² +x⁴)]
= 2[x¹⁰ - 10x⁸ + 15x⁶ - 10x⁴ + 5x²]

The given series is 1 + 5(-1)+9(1)+13(-1)+....
= 1 + 5(-1)+9(-1)²+13(-1)³+....
This is an arithmetic - geometric series with corresponding A.P. 1, 5, 9, 13.... and G.P. 1, -1, (-1)², (-1)³, ....
\(\therefore\) Tn of A.G. series = (Tn of A.P) (Tn of G.P)
= [ 1 + ( n - 1)4 ]  [ 1 (-1)^n-1]  [ \(\because\) T_n in A.P. is a + (n - 1)d, T_n in G.P. is a r^n-1 ]
= (4n-3)(-1)^n-1
Let Sn be the Sum of the first n terms of AG series
S_n = T₁ + T₂ + T₃ + ... T_n-1 + Tn
S_n = 1 + 5(-1)+9(-1)² + .. + (4n-7)(-1)^n-2 + (4n-3) (-1)^n-1  ...(2)
Multiplying by (-1) we get,
-1 S_n = 1 + 5(-1)+9(-3)³ + .. + (4n-7)(-1)^n-1 + 4n-3(-1)ⁿ
(2)-(3) we get,
S_n + S_n = 1 + [4(-1) + 4(-1)²]+...+4(-1)^n-1]-(4n-3)(-1)ⁿ
\(\Rightarrow\)   \(2S_n=1+{4(-1)[1-(-1)^{n-1}]\over1-(-1)}-(4n-3)(-1)^n\)    \(\begin{bmatrix} \because For\ GP, S_N={a(1-r^n)\over 1-r} \\a=1,r=-1 \end{bmatrix}\)
\(\Rightarrow\)    \(2S_n=1-2()1-(-1)^{n-1}-(4n-3){(-1)}^{n}\)
\(\Rightarrow\)    \(S_n={1\over 2}-1+{(-1)}^{n-1}-{4n-3\over 2}{(-1)}^{n-1}(-1)\)
\(\Rightarrow\)   \(S_n={-{1\over 2}}+{2+4n-3\over 2}{(-1)}^{n-1}\)
\(\Rightarrow\)   \(S_n={1\over 2}[-1+(4n-14){(--1)}^{n-1}].\)

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.

\(\frac { { \left( 1-2x \right) }^{ \frac { 2 }{ 3 } }{ \left( 4+5x \right) }^{ \frac { 3 }{ 2 } } }{ \sqrt { 1-x } } =\frac { \left( 1+\frac { 2 }{ 3 } (-2x)+..... \right) { \left( 4 \right) }^{ \frac { 3 }{ 2 } }{ \left( 1+\frac { 5 }{ 4 } x \right) }^{ \frac { 3 }{ 2 } } }{ { \left( 1-x \right) }^{ \frac { 1 }{ 2 } } } \)  [using binomial theorem for rational index]
\(={ \left( 1-\frac { 4x }{ 3 } \right) (8) }{ \left( 1+\frac { 3 }{ 2 } \left( \frac { 5 }{ 4 } x \right) \right) }\left( 1-\frac { 1 }{ 2 } (-x) \right) \) [neglecting x², x³ terms....]
\(=8\left( 1-\frac { 4x }{ 3 } \right) \left( 1+\frac { 15x }{ 8 } \right) \left( 1+\frac { x }{ 2 } \right) \)
\(=8\left( 1-\frac { 4x }{ 3 } +\frac { 15x }{ 8 } \right) \left( 1+\frac { x }{ 2 } \right) \)
\(=8\left( 1+\frac { 13x }{ 24 } \right) \left( 1+\frac { x }{ 2 } \right) =8\left( 1+\frac { 13x }{ 24 } +\frac { x }{ 2 } \right) =8\left( 1+\frac { 25x }{ 24 } \right) \)
When x = 0.001, the value of \(\frac { { \left( 1-2x \right) }^{ \frac { 2 }{ 3 } }{ \left( 4+5x \right) }^{ \frac { 3 }{ 2 } } }{ \sqrt { 1-x } } \)
= 8 + \(\frac { 25 }{ 3 } \)(0.001) = 8.01 (upto 2 places )

Given a = 5
Since the particle gets doubled, the G.P will be 5, 10,20,40, ... 1,50,000
⇒ a.r^n-1 > 1,50,000
⇒ a.(2^n-1) > 1,50,000
⇒ 5(2^n-1) > 1,50,000
⇒ \(2^{n-1}>{1,50,000\over5}\)
⇒ 2^n-1 > 30,000
⇒ 2^n-1 > 2⁴ x 1875
⇒ \({2^{n-1}\over24}>1875\)
⇒ 2^n-5 > 1875
⇒ (n - 5) log 2 > log 1875
⇒ \(n-5>{log1875\over log2}\)
⇒ \(n-5>{3.2730\over 0.3010}\)
⇒ n-5 > 10.873
⇒ n > 10.873 + 5
⇒ n > 15.873
⇒ n = 15
Hence the 15^th day, the infectious Virus particles just grow over 1,50,000 units.

According to the given information, we have the following diagram.

Distance travelled to bring first ball = 24 + 24 = 2 \(\times\) 24 = 48 m
Distance travelled to bring second ball = 2 (24 + 4) = 2(28) = 56 m
Distance travelled to bring third ball = 2 (24 + 4 + 4) = 2(32) = 64 m
\(\therefore\) The series of distances are 48, 56, 64 ...
Here a = 48, d = 56 - 48 = 8 and n = 20.
To find the total distance that he run in bringing back all balls, we have to find the sum of 20 terms of the above series
\(\therefore\) \({ S }_{ 20 }=\frac { 20 }{ 2 } \left[ 2\left( 48 \right) +19\left( 8 \right) \right] \)
= 10[96 + 152]
= 10[248]
S₂₀ = 2480 m.

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 series is \(\sqrt { 3 } +\sqrt { 75 } +\sqrt { 243 } +.... .\) and \(S_n =435\sqrt { 3 }\)
Given series is \(1(\sqrt3)+5(\sqrt3)+9(\sqrt3)+...\)
Here a = √3, d = 4√3
∴ The given series an arithmetic progression
\(∴\ S_n={n\over2}[2a+(n-1)d]\)
\(435\sqrt3={n\over2}[2\sqrt3 +(n -1)4\sqrt3]\)  [∵ given S_n = 435√3J]
\(435\sqrt3={n\over2}[2\sqrt3+4n\sqrt3-4\sqrt4]\)

\(⇒\ 435\sqrt3={n\over2}[4n\sqrt3-2\sqrt3]\)
\(⇒\ 435\sqrt3=2{\sqrt3.n\over2}[2n-1]\)
⇒ 435 = 2n²-n
⇒ 2n²- n - 435 = 0
⇒ (n = 15)(2n + 29) = 0
⇒ \(n-15\ or\ n={-29\over2}\) which is not possible
⇒ n = 15

Let T_n be the nth term of the given series
Then T_n = 1 + 4 + 4² + 4³ + ...
\(=1\left(4^n-1\over 4-1\right)\)
\(={4^n-1\over 3}\)
Let S_n be the sum to n terms of the given series
Then \(S_n={\sum_{k=1}^n}T_k=\sum_{k=1}^n{4^n-3\over3}\)
\(⇒\ S_n={1\over3}\left[ \sum_{k=1}^n4^n-\sum_{k=1}^n3\right]\)
\(⇒\ S_n= {1\over3}[4^1+4^]+...+4^n-3^n\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]={1\over 3}\left[4{(4^n-1)-9n\over3}\right]\)

\({ S }_{ n }=\frac { 4 }{ 9 } \left[ \left( { 4 }^{ n }-1 \right) -n/3 \right] \)

Let = 6 + 66 + 666 + ... upto n terms
= 6 (I + 11 + 111+ ....) upto n terms
\(={6\over9}(9+99+999+ ...)\) upto n terms
\(={63\over 6}[(10 -1) + (10^2-1) + (10^3 -1) + ...]\) upto n terms
\(={6\over 9}[(10+ 10^2 + 10^3+ ...) - (1+ 1+1...)]\) upto n terms​​​​​​​
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]\)[In a G.P with a = 10 r = 10, \(S_n={(r^n-1)\over r-1}\)]
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]={6\over9}\left[ 10(10^n-1)-9n\over9\right]\)
\({ S }_{ n }=\frac { 6 }{ 81 } \left[ 10\left( { 10 }^{ n }-1 \right) -9n \right] \)

Let S_n = 8 + 88 + 888 + 8888 + .... upto n terms
= 8 (1 + 11 + 111 + 1111 + ....) upto n terms
\(={8\over9}(9 + 99 + 999 + ...)\)
\({ S }_{ n }=\frac { 8 }{ 81 } \left[ \left( { 10 }^{ n }-1 \right) -9n \right] \) [multiplying and dividing by 9]
\(={8\over9}[10 -1) + (100 -1) + (1000 -1) + ...]\)
\(S_n={8\over 9}[(10^1 +10^2 +10^3 + ... +10^n)-(1+1+1+ ... +1n\ terms)]\)
In 10 + 10² + 10³ + ... + 10ⁿ, a = 10, r= 10, and it forms a G.P.
\(∴\ S_n={a(r^n-1)\over r-1}=10{(10^n-1)\over 10-1}={10\over 9}(10^n)-1\) and 1 + 1 + 1 ... + upto n terms = n
Substituting these values in (1) we get
\(S_n={8\over 9}\left[ 10(10^n-1)n\over 9\right]\)
\(S_n={8\over 81}[(10^n-1)-9n]\)

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
\(S_n=\sum_{k=1}^n{3k-2\over 3^{k-1}}\)
\(={\sum_{k=1}^n3k-2.{1\over{\sum_{k=1}^n}3^{k-1}}}\)
\(= 3[1+ 2 + 3+ ...+ n] - 2n \left[ 1\over3^0+3^2+...+3^{r-1}\right]\)
\(=\left[ 3{n(n+1)\over2}-2n\right]\left[ 1\over 1\left(3^n-1\over 3-1\right)\right]\)
 \(=\left[{3n^2+3n\over2}-2n\right]\left[2\over 3^n-1\right]={3n^2+3n-4n\over2}\times{2\over3^n-1}\)
\(\frac { { 3n }^{ 2 }-n }{ { 3 }^{ n }-1 } =\frac { n\left( n-1 \right) }{ { 3 }^{ n }-1 } \)

T_n = a + (n - 1)d
T_m+n = a + (m + n - 1)d
& T_m-n = a + (m - n - 1)d
T_m+n + T_m-n = a + (m + n - 1)d + a + (m - n - 1)d
= 2a + d(m + n - 1 + m - n - 1)
= 2a + d(2m - 2)
= 2[a + (m - 1)d]
T_m+n + T_m-n = 2. T_m