\(\mathrm{AM} \geq \mathrm{GM} \)
\(\Rightarrow a \geq g \)

\(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32} \ldots . . =\sqrt{2}+2 \sqrt{2}+3 \sqrt{2}+4 \sqrt{2} . \)
\(=\sqrt{2}[1+2+3+\ldots .] \)
\(S_{n} =\frac{\sqrt{2}[n(n+1)]}{2} \)
\(=\frac{n(n+1)}{\sqrt{2}} \)

\(n^{\text {th }} \text { term }=1-\frac{1}{2^{n}}=1-2^{-n}\)

\(\mathrm{AM}=16, \quad \mathrm{GM}=8, \quad \mathrm{HM}=?\)
\(\frac{a+b}{2}=16 \Rightarrow a+b=32\)
\(\sqrt{a b}=8 \Rightarrow a b=64\)
\(\therefore \mathrm{HM} =\frac{2 a b}{a+b} \)
\(=\frac{2(64)}{32}= 4 \)

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 t_k denote the k^th term of the given series.
Then \(t_k{1\over k(k+1)}.\) 
By using partial fraction we get
\({1 \over k(k+1)}={1\over k}-{1\over{k+1}}\)
Thus \(t_1+t_2+....+t_n=\left( 1-{1\over 2} \right)+\left( {1\over 2}+{1\over 3} \right)+\left( {1\over 3}-{1\over 4} \right)+...+\left( {1\over n} -{1\over n+1} \right)=1-{1\over n+1}.\)

Since \(\alpha ,\beta \) are the roots of the equation \(​​{ x }^{ 2 }-px+q=0\),we have
\(\alpha +\beta =-\frac { \left( -p \right) }{ 1 } =p\quad and\quad \alpha \beta =\frac { q }{ 1 } q\)
\(\therefore \log { (1+px+q{ x }^{ 2 }) } =\log { \left[ 1+(\alpha +\beta )x+\alpha \beta { x }^{ 2 } \right] } \)
\(=\log { \left[ (1+\alpha x)(1+\beta x) \right] } \)
\(=\log { (1+\alpha x)+log(1+\beta x) } \)
\(=\left( \alpha x-\frac { { \alpha }^{ 2 } }{ 2 } { x }^{ 2 }+\frac { { \alpha }^{ 3 } }{ 2 } { x }^{ 3 }+......+\infty \right) +\left( \beta x-\frac { { \beta }^{ 2 }{ x }^{ 2 } }{ 2 } +\frac { { \beta }^{ 3 }{ x }^{ 3 } }{ 3 } +.....\infty \right) \)
\(=(\alpha +\beta )x-\frac { { (\alpha }^{ 2 }+\beta ^{ 2 }) }{ 2 } { x }^{ 2 }+\frac { { (\alpha }^{ 3 }+\beta ^{ 3 }) }{ 3 } { x }^{ 3 }-...\infty \)
Hence proved. 

Let p = q+h
h is numerically very small and so h₁² h³ ... may be neglected
RHS \(={(n+1)p+(n-1)q\over (n-1)p+(n+1)q}={(n+1)(q+h)+(n-1)q\over (n+1)(q+h)+(n+1)q}\)
\(={nq+q-nh+h+nq-q\over nq-q+nh-h+nq+q}={2nq+(n+1)h\over2nq+(n-1)h}\)
\(={1+{n+1\over2n}.{q\over h}\over1+{n-1\over2n}.{h\over q}}=\left(1+{n+1\over2n}.{h\over q}\right)\left(1+{n-1\over2n}.{h\over q}\right)^{-1}\)
\(=\left({1+{n+1\over2n}.{h\over q}}\right)\left(1-{n-1\over2n}.{h\over q}\right)=1+\left({n+1\over 2n}-{n-1\over2n}\right){h\over q}\)
\(={1+{1\over n}}.{h\over q}\)
LHS \(={p\over q}^{1\over n}=\left(q+h\over q\right)^{1\over n}=\left(1+{h\over q}\right)^{1\over n}=1+{1\over n}.{h\over q}\)
From (1) and (2), LHS = RHS
Now \(\sqrt[8]{15\over16}={(8+1)(15)+(8-1)(16)\over(8-1)(15)+(8+1)(16)}\)  [n = 8, p = 15 and q = 16]
 \(={(9)(15)+7(16)\over(7)(15)+9(16)}={135+112\over105+144}={247\over 249}\)
\(\left( \frac { 15 }{ 16 } \right) =0.9919\)

\(\frac { { \left( 1-2x \right) }^{ \frac { 2 }{ 3 } }{ \left( 4+5x \right) }^{ \frac { 3 }{ 2 } } }{ \sqrt { 1-x } } =\frac { \left( 1+\frac { 2 }{ 3 } (-2x)+..... \right) { \left( 4 \right) }^{ \frac { 3 }{ 2 } }{ \left( 1+\frac { 5 }{ 4 } x \right) }^{ \frac { 3 }{ 2 } } }{ { \left( 1-x \right) }^{ \frac { 1 }{ 2 } } } \)  [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 )

We know (1 +x)n = nC₀ +nC₁ x +nC₂ x² +...+nC_n-1 x^n-1 +nC_nXⁿ 
Putting x = 8
(1 +8)ⁿ = nC₀ + nC₁ (8) + nC₂ (64) +..+ nC_n-1 8^n-1 + nC_n . 8N
9ⁿ = 1 + 8n + nC² (64) + nC₃ + ..+ nC_n 8^n-2
9ⁿ - 8n -1 is divisible by 64 for all positive integer n
putting N = n +1 we get
9ⁿ⁺¹ -8(n+1) -1 is divisible by 64 for all positive integer n
(9^n-1 -8n -8 -1 ) is divisible by 64
9^n-1 - 8n -9 is always divisible by 4

Given \(\left(x^2+{1\over x}\right)^5(1+x^3)^{50}\)
Let us expand \(\left(x^2+{1\over x}\right)^5\)
\(=(x^2)^5+5C_1(x^2)^4\left(1\over x\right)^1+5C_2(x^2)^3\left(1\over x\right)^2+5C_3(x^2)^2\left(1\over x\right)^3+5C_4(x^2)^1\left(1\over x\right)^4+{1\over x^5}\)
\(=x^{10}+5{x^8\over x}+{5\times4\over 2\times1}.{x^6\over x^2}+{5\times4\over 2\times1}{x^4\over x^3}+5{x^2\over x^4}+{1\over x^5}\)
\(=x^{10}+5x^7+10x^4+10x+{5\over x^2}+{11\over x^5}\)
General term in (1 + x³)⁵⁰
T_r+1 = 50C_r(1)^50-r (x³y)^r = 50C_r . x^3r
\(=\left(x^2+{1\over x}\right)^5(1+ x^3)^{30}\)
\(=\left(x^{10}+5x^7+10x^4+10x+{5\over x^2}+{1\over x^5}\right)(1+x^3)^{50}\)
\(=\left(x^{10}+5x^7+10x^4+10x+{5\over x^2}+{1\over x^5}\right)(50C_0+50C_1x^3+50C_2x^6+50C_3x^6+50C_3x^9+ ...)\)
\(=\left(x^{10}+5x^7+10x^4+10x+{5\over x^2}+{1\over x^5}\right)(1 + 50x^3 + 1225 x^6 + 19600 x^9 + ...)\)
Now, Co-efficient of x⁴
= 10[Constant term in (1 + x³)⁵⁰] + 10[Co-efficient of x³ in (1 + x³)⁵⁰] + 5[Co-efficient of x⁶] +1[Co-efficient of x⁹]
= 10(1) + 10(50C₁) + 5(50C₂) + 1(50C₃) [using (1)]
\(= 10+ 10 \times 50+ 5 \times{50\times49\over 2\times1}+{50\times49\times48\over 3\times2\times1}=10 + 500 + 6125 + 19600 = 26235\)
∴ Co-efficient of x⁴ is 26235

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.

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.

Let A be the first term and R be the common ratio of the given G.P.
Then a = p^th term ⇒ a = ARP^p-1
⇒ log a log A+(p-1 )logR...(1)
b = q^th term b = AR^q-1
⇒ log b = logA +(q-1) log R...(2)
c = r^th term ⇒ c = AR^r-1
⇒ log c = log A + (r-1) log R
Now, LHS = (q - r) log a + (r - p) log b + (p - q) log c
= (q - r) [log A + (p - 1) log R] + (r - p) [log A + (q - 1) log R] + (p - q)[log A + (r-1) log R]
= log A [q - r + r - p + P - q] + log R [(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q)]
= log A (0) + log R [pq - pr - q + r + qr - pq - r + p + rp - rq - p + q]
= log R [0] = 0
∴ (q - r) log a + (r- p) log b + (p - q) log c = 0.

Let us take a = 3 and b = 2x in the binomial expansion of (a + b)¹⁰.
Then, x⁶ will appear in the term containing (2x)⁶ and nowhere else. So the term containing x⁶ is
\(^{10}{C}_{4}a^4b^6={10\times 9\times8\times 7\over4\times 3\times 2\times 1 }3^4{(2x)}^{6}=210\times3^4\times2^6x^6\)
So coefficient of x³ in the expansion of (3 + 2x)¹⁰ is 210 \(\times\) 3⁴2⁶

Let the number be a and b
∴ a + b = \(\frac { 13 }{ 6 } \)
Let A₁, A₂,..A_2n be the 2n A,M s between a and b ....(1)
= \(2n\left( \frac { a+b }{ 2 } \right) =n(a+b)=n\times \frac { 13 }{ 6 } (2)\)   using (1)
Also  A₁+ A₂+ A_2n = 2n+1(given)   ...(3)
From (2)  and (3),\(\frac { 13n }{ 6 } \) = 2n+1
⇒   13n = 12n + 6
⇒    n = 6
∴  No of A.M's inserted = 2n - 2(6) = 12

\(\frac { 2 }{ 5 } +\frac { 2 }{ { 3.5 }^{ 3 } } +\frac { 2 }{ { 5.5 }^{ 5 } } ....\infty \)
\(=2\left[ \frac { 1 }{ 5 } +\frac { 1 }{ { 3.5 }^{ 3 } } +\frac { 1 }{ { 5.5 }^{ 5 } } ....\infty \right] \)
= \(2\left[ \left( \frac { 1 }{ 5 } \right) +\frac { \left( \frac { 1 }{ 5 } \right) ^{ 3 } }{ 3 } +\frac { \left( \frac { 1 }{ 5 } \right) ^{ 5 } }{ 3 } +..\infty \right] =2.\frac { 1 }{ 2 } log\left[ \frac { 1+\frac { 1 }{ 5 } }{ 1-\frac { 1 }{ 5 } } \right] \)\(\left[ \because \frac { 1 }{ 2 } log\left( \frac { 1+x }{ 1-x } \right) =x+\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 5 } }{ 5 } +\frac { { x }^{ 7 } }{ 7 } +.... \right] \) 
= \(log\left( \frac { \frac { 6 }{ 5 } }{ \frac { 4 }{ 5 } } \right) =log\left( \frac { 3 }{ 2 } \right) \)

\((1+x)^{ \frac { p }{ q } }=\left[ 1+\frac { p }{ q } x+\frac { \frac { p }{ q } \left( \frac { p }{ q } -1 \right) }{ 2! } { x }^{ 2 }+\frac { \left( \frac { p }{ q } \right) \left( \frac { p }{ q } -1 \right) \left( \frac { p }{ q } -2 \right) }{ 3! } { x }^{ 2 }+... \right] \)
= \({ 5 }^{ \frac { 2 }{ 3 } }\left( 1+\frac { { x }^{ 2 } }{ 5 } \right) ^{ \frac { 2 }{ 3 } }\)
= \({ 5 }^{ \frac { 2 }{ 3 } }\left( 1+\frac { 2 }{ 3 } \left( \frac { { x }^{ 2 } }{ 5 } \right) +\frac { \left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 3 } -1 \right) }{ 2! } \left( \frac { { x }^{ 2 } }{ 5 } \right) ^{ 2 }+\frac { \left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 3 } -1 \right) \left( \frac { 2 }{ 3 } -2 \right) }{ 3! } \left( \frac { { x }^{ 2 } }{ 5 } \right) ^{ 3 }+... \right) \)
= \({ 5 }^{ \frac { 2 }{ 3 } }\left[ 1+\frac { 2x^{ 2 } }{ 15 } +\left( \frac { 2 }{ 3 } \right) \left( \frac { -1 }{ 3 } \right) .\frac { { x }^{ 4 } }{ 50 } +\left( \frac { 2 }{ 3 } \right) \left( \frac { -1 }{ 3 } \right) \left( \frac { -2 }{ 3 } \right) \frac { { x }^{ 6 } }{ 125\times 6 } +.... \right] \)
= \({ 5 }^{ \frac { 2 }{ 3 } }\left[ 1+\frac { 2x^{ 2 } }{ 5 } -\frac { 2x^{ 2 } }{ 9\times 50 } +\frac { 4 }{ 3\times 3\times 3 } \frac { { x }^{ 6 } }{ 125\times 6 } .... \right] \)
= \({ 5 }^{ \frac { 2 }{ 3 } }\left[ 1+\frac { 2x^{ 2 } }{ 5 } -\frac { { x }^{ 4 } }{ 225 } +\frac { 2x^{ 6 } }{ 10125 } ..... \right] \).
The expansion is valid only if \(\left| \frac { { x }^{ 2 } }{ 5 } \right| <1\Rightarrow { x }^{ 2 }<5\)

(a+b)ⁿ = nC₀ aⁿb⁰ + nC₁ a^n-1 b₂ + nC₂ a^n-2 b² +... +nC_n a⁰ bⁿ, n \(\epsilon \) N
= 100⁴ + 4C₁ (100)³ (2)¹ + 4C₂ (100)²(2)² + 4C₃ (100)¹(2³) + 2⁴
=100000000 + 4(1000000)(2) + \(\frac { 4\times (3) }{ 2(1) } \) (10000)(4) + 400 (8) + 16
= 100000000 + 8000000 + 240000 + 3200 + 16 
= 108,243,216

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

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.

Given series is  1² + 3² + 5² +...
Let Tn be the nth term
T_n = (nth term of 1, 3, 5,...)²
= [1+(n-1)2]² = (1 + 2n - 2)² = (2n-1)²
= 4n² + 1 - 4n
∴ Sum of n terms =  \(\sum { 4{ n }^{ 2 } } -4n+1=4\sum { n^{ 2 } } -4\sum { n } +n\)
= \(4\frac { (n)(n+1)(2n+1) }{ 6 } -\frac { 4n(n+1)+n }{ 2 } \)
= \(\frac { n }{ 2 } \) [2(n + 1)(n + 1) - 6(n + 1) + 3]
= \(\frac { n(4{ n }^{ 2 }-1) }{ 3 } \)

odd terms are 6, 4, 2, 0...
t_n = 6 +( n -1 ) (-2) = 6 -2n + 2
= 8 -2n
Even terms are 10, 12, 14 , 16
Here a = 1 , d = 2
t_n = 10 + ( n - 1) (2) = 10 + 2n -2
= 8 + 2n
n^th term of the given sequence is  \(\begin{cases} 8-2n \\ 8+2n \end{cases}\)

\({ a }_{ n }=\begin{cases} n+1\quad if\quad n\quad is\quad odd \\ n\quad \quad if\quad n\quad is\quad even \end{cases}\)
a₁ = 1 + 1 = 2, a2 = 2, a₃ = 3 + 1 = 4
a₄ = 4, a₅ = 5 +1 = 6, a₆ = 6
hence the first 6 terms are 4, 2, 2, 4, 6, 6...