Sequences and Series

Note :

Sum of square of first n odd terms

1² + 3² + 5² + . . . . .+ n² = ${{n\left( {2n - 1} \right)\left( {2n + 1} \right)} \over 3}$

Given,

1² + 2. 2² + 3² + 2.4² + 5² + 2.6² + . . . . . .

A = Sum of first 20 terms

$\therefore\,\,\,$A = 1² + 2.2² + 3² + 24² + 5² + 2.6² + . . . . . .20 terms

Arrange those terms this way,

A = [1² + 3² + 5² + . . . . . 10 terms] + [ 2.2² + 2.4² + 2.6² + . . . . 10 terms]

A = [ 1² + 3² + 5² + . . . . 10 terms ] + 2.² [ 1² + 2² + 3² + . . . .10 terms ]

A = $ {{10 \times \left( {2.10 - 1} \right)\left( {2.10 + 1} \right)} \over 3} + {2.2^2}\left[ {{{10 \times 11 \times 21} \over 6}} \right]$

A = $ {{10 \times 19 \times 21} \over 3} + 8 \times {{10 \times 11 \times 21} \over 6}$

A =70 $ \times $ 19 + 70 $ \times $ 44

A = 70 $ \times $ 63

B = Sum of first 40 terms

Arrange those terms this way.

B = [1²+ 3² + 5² +. . . . 20 terms ] + [2.2² + 2.4² +. . . . . 20 terms ]

B = [1² + 3² + 5² + . . . . 20 terms] + 2.2² [1² + 2² + . . . 20 terms ]

B = ${{20 \times 39 \times 41} \over 3} + \,\,8\,\, \times {{20 \times 21 \times 41} \over 6}$

B = 260 $ \times $ 41 + 560 $ \times $ 41

B = 41 $ \times \,\,\,820$

$\therefore\,\,\,$ B $-$ 2A = 41 $ \times \,$ 820 $-$ 2 $ \times \,$ 70 $ \times \,$ 63 = 24800

Given that B $-$ 2A = 100 $\lambda $

$\therefore\,\,\,$ 100 $\lambda $ = 24800

$ \Rightarrow \,\,\,\lambda $ = 248

a₁, a₂, a₃ . . . a₄₃ are in AP

So, a₂ = a₁ + d

a₃ = a₁ + 2d

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a₄₉ =a₁ + 48d

Now given, ${a_9} + {a_{43}} = 66$

$ \Rightarrow \,\,\,\,$ a₁ + 8d + a₁ + 42d = 66

$ \Rightarrow \,\,\,\,$ 2a₁ + 50d = 66

$ \Rightarrow \,\,\,\,$ a₁ + 25d = 33 . . . . . (1)

$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} $ = 416

$ \Rightarrow \,\,\,\,$ a₁ + a₅ + a₉ + a₁₃ +. . . . . 13 items = 416

$ \Rightarrow \,\,\,\,$ a₁ + a₁ + 4d + a₁ + 8d + . . . . a₁ + 48d = 416

$ \Rightarrow \,\,\,\,$ 13a₁ + 4d +8d + 12d + . . . . . 48d = 416

$ \Rightarrow \,\,\,\,$ 13a₁ + 4 (1+ 2 + 3 + . . . + 12) d = 416

$ \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times $d = 416

$ \Rightarrow \,\,\,\,$ 13a₁ + 24 $ \times$ 13d = 416

$ \Rightarrow \,\,\,\,$ a₁ + 24 d =32 . . . .(2)

Solving (1) and (2) we get,

d = 1

and ${a_1} = 8$

$\therefore\,\,\,$ a₁ = 8

a₂ = 8 + 1 = 9

a₃ = 8 + 2 = 10

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a₁₇ = 8 + 16 = 24

Now, $a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$

$ \Rightarrow \,\,\,\,$ $a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$

$ \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$

We can write above series like this,

$ \Rightarrow \,\,\,\,\,$ (1² +2² + . . . . +24²) $-$ (1² + 2² + . . . . .+ 7²) = 140 m

$ \Rightarrow {{24\left( {25} \right)\left( {49} \right)} \over 6} - {{7 \times 8 \times 15} \over 6} = 140\,m$

$ \Rightarrow \,\,\,\,\,$ 490 $-$ 140 = 140 m

$ \Rightarrow \,\,\,\,\,$4760 = 140 m

$ \Rightarrow \,\,\,\,\,$ m = 34

A_n = $\left( {{3 \over 4}} \right) - {\left( {{3 \over 4}} \right)^2} + {\left( {{3 \over 4}} \right)^3} - .... + {\left( { - 1} \right)^{n - 1}}{\left( {{3 \over 4}} \right)^n}$

Which in a G.P. with a = ${{3 \over 4}}$, r = ${{{ - 3} \over 4}}$ and number of terms = n

$\therefore\,\,\,$ A_n = ${{{3 \over 4}\left( {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right)} \over {1 - \left( {{{ - 3} \over 4}} \right)}} = {{{3 \over 4} \times \left( {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right)} \over {{7 \over 4}}}$

$ \Rightarrow $$\,\,\,$A_n = ${{3 \over 7}}$$\left[ {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right]$ $\,\,\,\,\,\,\,\,\,$ . . . . . . . . . .(1)

As, B_n = 1 $-$ A_n

For least odd natural number p, such that B_n > A_n

$ \Rightarrow $$\,\,\,$ 1 $-$ A_n > A_n $ \Rightarrow $ 1 > 2 $ \times $ A_n $ \Rightarrow $ A_n < ${{1 \over 2}}$

From eqn. (1), we get

${{3 \over 7}}$$ \times $ $\left[ {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right] < {1 \over 2}$ $ \Rightarrow $ 1 $-$ ${\left( {{{ - 3} \over 4}} \right)^n} < {7 \over 6}$

$ \Rightarrow $$\,\,\,$ 1 $-$ ${7 \over 6}$ < ${\left( {{{ - 3} \over 4}} \right)^n}$   $ \Rightarrow $   ${{ - 1} \over 6} < {\left( {{{ - 3} \over 4}} \right)^n}$

As n is odd, then ${\left( {{{ - 3} \over 4}} \right)^n}$ = $-$ ${{{{3^n}} \over 4}}$

So ${{ - 1} \over 6}$ < $-$ ${\left( {{3 \over 4}} \right)^n}$ $ \Rightarrow $ ${1 \over 6}$ > ${\left( {{3 \over 4}} \right)^n}$

log$\left( {{1 \over 6}} \right)$ = n log$\left( {{3 \over 4}} \right)$ $ \Rightarrow $ 6.228 < n

Hence, n should be 7.

$ \because $$\,\,\,$a, b, c are in A.P. then

a + c = 2b

also it is given that,

a + b + c = ${{3 \over 4}}$                          . . . .(1)

$ \Rightarrow $$\,\,\,$ 2b + b = ${{3 \over 4}}$   $ \Rightarrow $$\,\,\,$ b = ${{1 \over 4}}$  . . . . .(2)

Again it is given that, a², b², c² are in G.P. then

(b²)² = a²c²   $ \Rightarrow $  ac = $ \pm $ ${{1 \over {16}}}$      . . . . (3)

From (1), (2) and (3), we get;

$a \pm {1 \over {16a}}$ = ${1 \over 2}$  $ \Rightarrow $ 16a² $-$ 8a $ \pm $ 1 = 0

Case I : 16a² $-$ 8a + 1 = 0

$ \Rightarrow $$\,\,\,$a = ${1 \over 4}$ (not possible as a < b)

Case II: 16a² $-$ 8a $-$ 1 = 0 

$ \Rightarrow $$\,\,\,$ a = ${{8 \pm \sqrt {128} } \over {32}}$

$ \Rightarrow $$\,\,\,$ a = ${1 \over 4} \pm {1 \over {2\sqrt 2 }}$

$ \therefore $$\,\,\,$ a = ${1 \over 4} - {1 \over {2\sqrt 2 }}$   ($ \because $ a < b)

Assume d₁ is the common difference of A.P x₁,x₂ ..... x_n

Given x₃ = 8 and x₈ = 20

$ \therefore $ x₁ + 2d₁ = 8 ..... (i)
and x₁ + 7d₁ = 20 ..... (ii)

Solving (i) and (ii) we get x₁ = $16 \over {15}$ and d₁ = $12 \over {5}$

Now let $1 \over d_2$ is the common difference of A.P $1 \over h_1$, $1 \over h_2$ ..... $1 \over h_n$

Given that,
h₂ = 8 and h₇ = 20

$ \therefore $ $1 \over h_2$ = $1 \over 8$

$ \Rightarrow $ $1 \over h_1$ + $1 \over d_2$ = $1 \over 8$ .... (iii)

and $1 \over h_7$ = $1 \over 20$

$ \Rightarrow $ $1 \over h_1$ + $6 \over d_2$ = $1 \over 20$ ... (iv)

Solving (iii) and (iv) we get
$1 \over h_1$ = $28 \over 200$ and $1 \over d_2$ = $- {3 \over 200}$
So, x₅ = x₁ + 4d₁
= $16 \over 5$ + $48 \over 5$= $64 \over 5$ and
$1 \over h_{10}$ = $1 \over h_1$ + $9 \over d_2$
= $28 \over 200$ - $27 \over 200$ = $1\over 200$

$ \therefore $ x₅ $\times$ h₁₀ = ${64 \over 5} \times 200$ = 2560

Sum of infinite G.P,

S = $b \over {1-r}$ where $\left| r \right| < 1$

$ \Rightarrow $ 5 = $b \over {1-r}$

$ \Rightarrow $ 1 - r = $b \over 5$

$ \Rightarrow $ b = 5(1 - r)

as $\left| r \right| < 1$

$ \therefore $  -1 < r < 1

$ \Rightarrow $ 1 > -r > -1

$ \Rightarrow $ 2 > 1-r > 0

$ \Rightarrow $ 10 > 5(1-r) > 0

$ \Rightarrow $ 10 > b > 0

$ \therefore $ interval of b = (0, 10)

n^th term, T_n = ${{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}$

T_n = ${{{{n\left( {n + 1} \right)} \over 2}} \over {{{\left( {{{n\left( {n + 1} \right)} \over 2}} \right)}^2}}}$

$ \Rightarrow $ T_n = ${2 \over {n\left( {n + 1} \right)}}$ = $2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$

$ \therefore $ S_n = $\sum {{T_n}} $

= $2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]} $

= $2\left( {1 - {1 \over n}} \right)$

= ${{{2n} \over {n + 1}}}$

Given that,

100 S_n = n

$ \Rightarrow $ 100 $ \times $ ${{{2n} \over {n + 1}}}$ = n

$ \Rightarrow $ n + 1 = 200

$ \Rightarrow $ n = 199

a, b and c are in AP.

$ \therefore $ a + c = 2b

As, abc = 8

$ \Rightarrow $ac$\left( {{{a + c} \over 2}} \right)$= 8

$ \Rightarrow $ ac(a + c) = 16 = 4 $ \times $ 4

$ \therefore $ ac = 4 and a + c = 4

Then,

b = $\left( {{{a + c} \over 2}} \right)$ = ${4 \over 2}$ = 2

Given,

$\sqrt 3 $ + $\sqrt {75} $ + $\sqrt {243} $ + $\sqrt {507} $ + . . . . . .+ n terms

= $\sqrt 3 $ + $\sqrt {25 \times 3} $ + $\sqrt {81 \times 3} $ + $\sqrt {169 \times 3} $ + . . . . . .+ n terms

= $\sqrt 3 $ + 5$\sqrt 3 $ + 9$\sqrt 3 $ + 13$\sqrt 3 $ + . . . . . .+ n terms

= $\sqrt 3 $ [ 1 + 5 + 9 + 13 + . . . . .+ n terms]

= $\sqrt 3 $ $\left[ {{n \over 2}\left( {2.1 + \left( {n - 1} \right)4} \right)} \right]$

= $\sqrt 3 $ $\left[ {{n \over 2}\left( {2 + 4n - 4} \right)} \right]$

= $\sqrt 3 $ $\left[ {{n \over 2}\left( {4n - 2} \right)} \right]$

= $\sqrt 3 $ [n (2n $-$ 1)]

According to question,

$\sqrt 3 $ [n (2n $-$ 1)] = 435$\sqrt 3 $

$ \Rightarrow $$\,\,\,$ 2n² $-$ n = 435

$\therefore\,\,\,$ n = ${{1 \pm \sqrt {1 + 4 \times 2 \times 435} } \over 4}$ = ${{1 \pm 59} \over 4}$

$\therefore\,\,\,$ n = ${{1 + 59} \over 4}$ = 15   or   ${{1 - 59} \over 4}$ = $-$ 14.5

$\therefore\,\,\,$ n = 15 (as    n  can't be $-$ve)

A.T.Q.,

A.M. = 5G.M.

${{a + b} \over 2} = 5\sqrt {ab} $

${{a + b} \over {\sqrt {ab} }}$ $ = 10$

$ \therefore $   ${a \over b} = {{10 + \sqrt {96} } \over {10 - \sqrt {96} }} = {{10 + 4\sqrt 6 } \over {10 - 4\sqrt 6 }}$

Use componendo and Dividendo

${{a + b} \over {a - b}} = {{20} \over {8\sqrt 6 }} = {5 \over {2\sqrt 6 }} = {{5\sqrt 6 } \over {12}}$

9(25${a^2}$ + b²) + 25(c² - 3$a$c) = 15b(3$a$ + c)

$ \Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc$

$ \Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + {\left( {5c} \right)^2} - 75ac = 45ab + 15bc$

$ \Rightarrow $ ${1 \over 2}\left[ {{{\left( {15a - 3b} \right)}^2} + {{\left( {3b - 5c} \right)}^2} + {{\left( {5c - 15a} \right)}^2}} \right] = 0$

it is possible when 15a – 3b = 0, 3b – 5 c = 0 and 5c – 15a = 0

$ \Rightarrow $ 15a = 3b = 5c

$ \Rightarrow $ b = ${{5c} \over 3}$, a = ${c \over 3}$

$ \Rightarrow $ a + b = ${c \over 3} + {{5c} \over 3}$ = ${{6c} \over 3}$ = 2c

$ \therefore $ b, c, a are in A.P.

Given,

A + B = ${\pi \over 6}$

$ \therefore $   tan(A + B) = tan$\left( {{\pi \over 6}} \right)$ = ${1 \over {\sqrt 3 }}$

We know,

tan(A + B) = ${{\tan A + \tan B} \over {1 - \tan A\tan B}}$

$ \Rightarrow $  ${1 \over {\sqrt 3 }}$ = ${y \over {1 - \tan A\tan B}}$

where y = tan A + tan B

$ \Rightarrow $    tanA tanB = 1 $-$ $\sqrt 3 $ y

Also AM   $ \ge $  GM

$ \Rightarrow $    ${{\tan A + \tan B} \over 2} \ge \sqrt {\tan A\tan B} $

$ \Rightarrow $   y $ \ge $ 2$\sqrt {1 - \sqrt 3 y} $

$ \Rightarrow $   y² $ \ge $ 4 $-$ 4${\sqrt 3 y}$

$ \Rightarrow $    y² + 4${\sqrt 3 y}$ $-$ 4 $ \ge $ 0

$ \Rightarrow $   y  $ \le $ $-$ 2$\sqrt 3 $ $-$ 4   

or   y $ \ge $ $-$ 2$\sqrt 3 $ + 4

(y $ \le $ $-$ 2$\sqrt 3 $ $-$ 4 is not possible as tan B > 0)

z = 1 + ai

z² = 1 $-$ a² + 2ai

z² . z = {(1 $-$ a²) + 2ai}   {1 + ai}

= (1 $-$ a²) + 2ai + (1 $-$ a²)    ai $-$ 2a²

$ \because $    z³ is real  $ \Rightarrow $  2a + (1 $-$ a²) a = 0

a (3 $-$ a²) = 0   $ \Rightarrow $  a = $\sqrt 3 $ (a > 0)

1 + z + z² . . . . . . . z¹¹ = ${{{z^{12}} - 1} \over {z - 1}} = {{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {1 + \sqrt 3 i - 1}}$

= ${{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {\sqrt 3 i}}$

(1 + ${\sqrt 3 i}$)¹² = 2¹² ${\left( {{1 \over 2} + {{\sqrt 3 } \over 2}i} \right)^{12}}$

= 2¹² (cos${\pi \over 3}$ + isin${\pi \over 3}$)¹² = 2¹² (cos4$\pi $ + isin4$\pi $) = 2¹²

$ \Rightarrow $    ${{{2^{12}} - 1} \over {\sqrt 3 i}} = {{4095} \over {\sqrt 3 i}} = - {{4095} \over 3}\sqrt 3 i = - 1365\sqrt 3 i$

As  a₁ a₂ . . . . . a_n . . . . . are in A.P.

$ \therefore $   a₃ + a₁₅ = a₇ + a₁₁ = a₁ + a₁₇

Given,

a₃ + a₇ + a₁₁ + a₁₅ + a₁₅ = 72

$ \Rightarrow $   (a₃ + a₁₅) + (a₇ + a₁₁) = 72

$ \Rightarrow $   2(a₁ + a₁₇) = 72

$ \Rightarrow $   (a₁ + a₁₇) = 36

$ \therefore $   Sum of first 17 terms

= ${{17} \over 2}$ (a₁ + a₁₇)

= ${{17} \over 2}$ $ \times $ 36

= 306

As we know

AM  $ \ge $  GM

$ \Rightarrow $   ${{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}$  $ \ge $  ${\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right)}^4}{{\left( {{z \over 5}} \right)}^5}} \right]^{{1 \over {12}}}}$

$ \Rightarrow $   1  $ \ge $  ${{{x^3}{y^4}{z^5}} \over {{3^3}{4^4}{5^5}}}$

$ \Rightarrow $   x³ y⁴ z⁵  $ \le $  3³ . 4⁴ . 5⁵

$ \Rightarrow $   x³ y⁴ z⁵  $ \le $  (0.1)(600)³

but given that,

x³ y⁴ z⁵ = (0.1) (600)³

$ \therefore $   AM  $=$  GM

$ \Rightarrow $   All the number are equal.

$ \therefore $   ${x \over 3} = {y \over 4} = {z \over 5} = k$

$ \Rightarrow $  x $=$ 3k, y = 4k, z = 5k

given that,

        x + y + z $=$ 12

$ \Rightarrow $   3k + 4k + 5k $=$ 12

$ \Rightarrow $   12k $=$ 12

$ \Rightarrow $   k = 1

$ \therefore $   x $=$   3,  y $=$ 4,   z $=$ 5

So, x³ + y³ + z³

$=$ 3³ + 4³ + 5³

$=$ 216

The terms of an Arithmetic Progression (A.P.) are given by $a$, $a + d$, $a + 2d$, ..., where $a$ is the first term and $d$ is the common difference.

Given that the 2nd, 5th and 9th terms of an A.P. are in Geometric Progression (G.P.), we can denote them as follows :

2nd term = $a + d$

5th term = $a + 4d$

9th term = $a + 8d$

For three numbers to be in G.P., the square of the middle term must be equal to the product of the other two terms. So,

$(a + 4d)^2 = (a + d)(a + 8d)$

Expanding and simplifying :

$a^2 + 8ad + 16d^2 = a^2 + 9ad + 8d^2$

$8ad + 16d^2 = a^2 + 9ad + 8d^2$

$8ad - 9ad = 8d^2 - 16d^2$

$-ad = -8d^2$

$a = 8d$

The common ratio of the G.P. is the ratio of the 5th term to the 2nd term, or $(a + 4d) / (a + d)$. Substituting $a = 8d$ gives :

$(8d + 4d) / (8d + d) = 12d / 9d = 4 / 3$

So, the common ratio of the G.P. is $4 / 3$. The answer is option D.

${\left( {{8 \over 5}} \right)^2} + {\left( {{{12} \over 5}} \right)^2} + {\left( {{{16} \over 5}} \right)^2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {{{20} \over 5}} \right)^2}.... + {\left( {{{44} \over 5}} \right)^2}$

$S = {{16} \over {25}}\left( {{2^2} + {3^2} + {4^2} + ...... + {{11}^2}} \right)$

$ = {{16} \over {25}}\left( {{{11\left( {11 + 1} \right)\left( {22 + 1} \right)} \over 6} - 1} \right)$

$ = {{16} \over {25}} \times 505 = {{16} \over 5} \times 101$

$ \Rightarrow {{16} \over 5}m = {{16} \over 5} \times 101$

$ \Rightarrow m = 101.$

${n^{th}}$ term of series

$ = {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$

Sum of $n$ term $ = \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$

$ = {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \right]$

$ = {1 \over 4}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6} + {{2n\left( {n + 1} \right)} \over 2} + n} \right]$

Sum of $9$ terms

$ = {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96$

$m = {{l + n} \over 2}$ and common ratio of

$G.P.$ $ = r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$

$\therefore$ ${G_1} = {l^{3/4}}\,{n^{1/4}},$ ${G_2} = {l^{1/2}}{n^{1/2}},\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$

$G_1^4 + 2G_2^4 + G_3^4$

$ = {l^3}n + 2{l^2}{n^2} + {\ln ^3}$

$ = \ln {\left( {1 + n} \right)^2}$

$ = \ln \times 2{m^2}$

$ = 4l{m^2}n$

Let $a,ar,a{r^2}$ are in $G.P.$

According to the question

$a,2ar,a{r^2}$ are in $A.P.$

$ \Rightarrow 2 \times 2ar = a + a{r^2}$

$ \Rightarrow 4r = 1 + {r^2}$

$ \Rightarrow {r^2} - 4r + 1 = 0$

$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $

Since $r > 1$

$\therefore$ $\pi = 2 - \sqrt 3 $ is rejected

Hence, $r = 2 + \sqrt 3 $

Let ${10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$

Let $x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}$

Multiplied by ${{11} \over {10}}$ on both the sides

${{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}$

$x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}$

$ \Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}$

$ \Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}$

$ \Rightarrow x = {10^{11}} = k{.10^9}$

Given $ \Rightarrow k = 100$

Given sequence can be written as

${7 \over {10}} + {{77} \over {100}} + {{777} \over {{{10}^3}}} + ..... + $ up to $20$ terms

$ = 7\left[ {{1 \over {10}} + {{11} \over {100}} + {{111} \over {{{10}^3}}} + ...... + } \right.\,\,$ up to $20$ terms ]

Multiply and divide by $9$

$ = {7 \over 9}\left[ {{9 \over {10}} + {{99} \over {100}} + {{999} \over {1000}} + ......} \right.\,\,$ $+$ up to $20$ terms ]

$ = {7 \over 9}\left[ {\left( {1 - {1 \over {10}}} \right)} \right. + \left( {1 - {1 \over {{{10}^2}}}} \right) + \left( {1 - {1 \over {{{10}^3}}}} \right) + ......$ $+$ up to $20$ terms ]

$ = {7 \over 9}\left[ {20 - {{{1 \over {10}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^{20}}} \right)} \over {1 - {1 \over {10}}}}} \right]$

$ = {7 \over 9}\left[ {{{179} \over 9} + {1 \over 9}{{\left( {{1 \over {10}}} \right)}^{20}}} \right]$

$ = {7 \over {81}}\left[ {179 + {{\left( {10} \right)}^{ - 20}}} \right]$

$n$^th term of the given series

$ = {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}$

$ = {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}$

$ = {n^3} - {\left( {n - 1} \right)^3}$

$ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left[ {{k^3} - {{\left( {k - 1} \right)}^3}} \right]} $

$ \Rightarrow 8000 = {n^3}$

$ \Rightarrow n = 20\,\,$ which is a natural number.

Now, put $n = 1,2,3,.....20$

${T_1} = {1^3} - {0^3}$

${T_2} = {2^3} - {1^3}$

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.

.

${T_{20}} = {20^3} - {19^3}$

Now, ${T_1} + {T_2} + ..... + {T_{20}} = {S_{20}}$

$ \Rightarrow {S_{20}} = {20^3} - {0^3} = 8000$

Hence, both the given statements are true and statement $2$ supports statement $1.$

Let required number of months $=n$

$\therefore$ $200 \times 3 + \left( {240 + 280 + 320 + ...} \right.$

$\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040$

$ \Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600$

$ \Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440$

$ \Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440$

$ \Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522$

$ \Rightarrow {n^2} + 5n - 546 = 0$

$ \Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0$

$\therefore$ $n = 21$

Till $10$^th minute number of counted notes $ = 1500$

$3000 = {n \over 2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = n\left[ {148 - n + 1} \right]$

$ \Rightarrow $${n^2} - 149n + 3000 = 0$

$ \Rightarrow n = 125,24$

But $n=125$ is not possible

$\therefore$ total time $ = 24 + 10 = 34$ minutes.

We have

$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........\infty \,\,\,\,\,...\left( 1 \right)$

Multiplying both sides by ${1 \over 3}$ we get

${1 \over 3}S = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + {{10} \over {{3^4}}} + .......\,\,\,\,\,...\left( 2 \right)$

Subtracting eqn. $(2)$ from eqn. $(1)$ we get

${2 \over 3}S = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty $

$ \Rightarrow {2 \over 3}S = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty $

$ \Rightarrow {2 \over 3}S = {{{4 \over 3}} \over {1 - {1 \over 3}}} = {4 \over 3} \times {3 \over 2}$

$ \Rightarrow S - 3$

As per question,

$\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

$ \Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} = {{48} \over {12}}$

$ \Rightarrow {r^2} = 4, \Rightarrow r = - 2$

(As terms are $=+ve$ and $-ve$ alternately)

$ \Rightarrow a = - 12$

Let the series $a,ar,$ $a{r^2},........$ are in geometric progression.

given, $a = ar + a{r^2}$

$ \Rightarrow 1 = r + {r^2}$

$ \Rightarrow {r^2} + r - 1 = 0$

$ \Rightarrow r = {{ - 1 \mp \sqrt {1 - 4 \times - 1} } \over 2}$

$ \Rightarrow r = {{ - 1 \pm \sqrt 5 } \over 2}$

$ \Rightarrow r = {{\sqrt 5 - 1} \over 2}$

[ As terms of $G.P.$ are positive

$\therefore$ $r$ should be positive]

We know that ${e^x} = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ........\infty $

Put $x=-1$

$\therefore$ ${e^{ - 1}} = 1 - 1 + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}..........\infty $

$\therefore$ ${e^{ - 1}} = {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - {1 \over {5!}}........\infty $

${1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d$ (say)

Then ${a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......,\,{a_{n - 1}}{a_n} = {{{a_{n - 1}} - {a_n}} \over d}$

$\therefore$ ${a_1}a{}_2 + {a_2}{a_3} + ......... + {a_{n - 1}}{a_n}$

$ = {{{a_1} - {a_2}} \over d} + {{{a_2} - {a_3}} \over d} + ......$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{{a_{n - 1}} - {a_n}} \over d}$

$ = {1 \over a}\left[ {{a_1}} \right. - {a_2} + {a_2} - {a_3} + .......$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + {a_{n - 1}} - an} \right] = {{{a_1} - {a_n}} \over d}$

Also, ${1 \over {{a_n}}} = {1 \over {{a_1}}} + \left( {n - 1} \right)d$

$ \Rightarrow {{{a_1} - {a_n}} \over {{a_1}{a_n}}} = \left( {n - 1} \right)d$

$ \Rightarrow {{{a_1} - {a_n}} \over d} = \left( {n - 1} \right){a_1}{a_n}$

Which is the required result.

${{{p \over 2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]} \over {{q \over 2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = {{{p^2}} \over {{q^2}}}$

$ \Rightarrow {{2{a_1} + \left( {p - 1} \right)d} \over {2{a_1} + \left( {p - 1} \right)d}} = {p \over q}$

${{{a_1} + \left( {{{p - 1} \over 2}} \right)d} \over {{a_1} + \left( {{{q - 1} \over 2}} \right)d}} = {p \over q}$

For ${{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41$

$ \Rightarrow {{{a_6}} \over {a{}_{21}}} = {{11} \over {41}}$

${{{e^x} + {e^{ - x}}} \over 2}$

$ = 1 + {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} + {{{x^6}} \over {6!}}.........$

Putting $x = {1 \over 2}$ we get

$1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......$

$\infty = {{{e^{{1 \over 2}}} + {e^{{{ - 1} \over 2}}}} \over 2} = {{\sqrt e + {1 \over {\sqrt e }}} \over 2}$

$ = {{e + 1} \over {2\sqrt e }}$

$x = \sum\limits_{n = 0}^\infty {{a^n}} = {1 \over {1 - a}}\,\,\,\,\,\,\,\,\,\,a = 1 - {1 \over x}$

$y = \sum\limits_{n = 0}^\infty {{b^n}} = {1 \over {1 - b}}\,\,\,\,\,\,\,\,\,\,b = 1 - {1 \over y}$

$z = \sum\limits_{n = 0}^\infty {{c^n}} = {1 \over {1 - c}}\,\,\,\,\,\,\,\,\,\,c = 1 - {1 \over z}$

$a,b,c$ are in $A.P.$ OR $2b = a + c$

$2\left( {1 - {1 \over y}} \right) = 1 - {1 \over x} + 1 - {1 \over y}$

${2 \over y} = {1 \over x} + {1 \over z} \Rightarrow x,y,z$ are in $H.P.$

We know that

$e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........$

and

${e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........$

$\therefore$ $e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \right]$

$\therefore$ ${1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + .......$

$ = {{e + {e^{ - 1}}} \over 2} - 1$

$ = {{{e^2} + 1 - 2e} \over {2e}}$

$ = {{{{\left( {e - 1} \right)}^2}} \over {2e}}$

${T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)$

${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$

$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$

$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$

From $\left( 1 \right)$ $a = {1 \over {mn}} \Rightarrow a - d = 0$

If $n$ is odd, the required sum is

${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$

$ = {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}$

[ As $\left( {n - 1} \right)$ is even

$\therefore$ using given formula for the sum of $\left( {n - 1} \right)$ terms.]

$ = \left( {{{n - 1} \over 2} + 1} \right){n^2} = {{{n^2}\left( {n + 1} \right)} \over 2}$

${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty $

$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$

$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $

$ = \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$

$ = 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$

$ = 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]$

$ = 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$

The product is $p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........$

$ = {2^{1/4 + 2/8 + 3/16 + .......\infty }}$

Now let

$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$

${1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,........\left( 2 \right)$

Subtracting $(2)$ from $(1)$

$ \Rightarrow {1 \over 2}S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + .......\infty $

or ${1 \over 2}S = {{1/4} \over {1 - 1/2}} = {1 \over 2} \Rightarrow S = 1$

$\therefore$ $P = {2^S} = 2$

${1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}$

$ = {1^3} + {2^3} + {3^3} + ...... + {9^3}$

$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$

$ = {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]$

$ = {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}$

$ = 2025 - 1600 = 425$

$l = A{R^{p - 1}}$

$ \Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$

$m = A{R^{q - 1}}$

$ \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$

$n = A{R^{r - 1}}$

$ \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$

Now, $\left| {\matrix{ {\log l} & p & 1 \cr {\log m} & q & 1 \cr {\log n} & r & 1 \cr } } \right|$

$ = \left| {\matrix{ {\log A + \left( {p - 1} \right)\log R} & p & 1 \cr {\log A + \left( {q - 1} \right)\log R} & q & 1 \cr {\log A + \left( {r - 1} \right)\log R} & r & 1 \cr } } \right|$

Operating ${C_1} - \left( {\log R} \right){C_2} + \left( {\log R - \log A} \right){C_3}$

$ = \left| {\matrix{ 0 & p & 1 \cr 0 & q & 1 \cr 0 & r & 1 \cr } } \right| = 0$

Let $a=$ first team of $G.P.$ and $r=$ common ratio of $G.P.;$

Then $G.P.$ is $a,$ $ar,$ $a{r^2}$

Given ${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$

$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$

Also ${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$ to $\infty = 100$

$ \Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100$

$ \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)$

From $(i),$ ${a^2} = 400{\left( {1 - r} \right)^2};$

From $(ii),$ we get $100\left( {1 - r} \right)\left( {1 + r} \right)$

$\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}$

$ \Rightarrow 1 + r = 4 - 4r$

$ \Rightarrow 5r = 3$

$ \Rightarrow r = 3/5.$

$1,\,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log _3}\left( {{{4.3}^x} - 1} \right)$ are in $A.P.$

$ \Rightarrow 2{\log _9}\left( {{3^{1 - x}} + 2} \right)$

$\,\,\,\,\,\,\,\,\,$ $ = 1 + {\log _3}\left( {{{4.3}^x} - 1} \right)$

$ \Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)$

$\,\,\,\,\,\,\,\,\,$ $ = {\log _3}3 + {\log _3}\left( {{{4.3}^x} - 1} \right)$

$ \Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)$

$\,\,\,\,\,\,\,\,\,$ $ = {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right]$

$ \Rightarrow {3^{1 - x}} + 2 = 3\,\left( {{{4.3}^x} - 1} \right)$

$ \Rightarrow {3.3^{ - x}} + 2 = {12.3^x} - 3.$

Put ${3^x} = t$

$ \Rightarrow {3 \over t} + 2 = 12t - 3$

or $12{t^2} - 5t - 3 = 0;$

Hence $t = - {1 \over 3},{3 \over 4} \Rightarrow {3^x} = {3 \over 4}$

(as ${3^x}\,\, \ne \,\, - ve$ )

$ \Rightarrow x = {\log _3}\left( {{3 \over 4}} \right)$

or $x = {\log _3}3 - {\log _3}4$

$ \Rightarrow x = 1 - {\log _3}4$

$a{r^4} = 2$

$a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$

$ = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512$