Sequences and Series

If logb₁, logb₂, ......, logb₁₀₁ are in A.P. with common difference log_e2, then b₁, b₂, ......, b₁₀₁ are in G.P., with common ratio 2.

$\therefore$ b₁ = 2⁰b₁

b₂ = 2¹b₁

b₃ = 2²b₁

$\matrix{ \vdots & \vdots & \vdots \cr } $

b₁₀₁ = 2¹⁰⁰b₁ ..... (i)

Also, a₁, a₂, ......, a₁₀₁ are in A.P.

Given, a₁ = b₁ and a₅₁ = b₅₁

$\Rightarrow$ a₁ = b₁ and a₅₁ = b₅₁

$\Rightarrow$ a₁ + 50D = 2⁵⁰ b₁

$\Rightarrow$ a₁ + 50D = 2⁵⁰ a₁ [$\because$ a₁ = b₁] ...... (ii)

Now, t = b₁ + b₂ + ..... + b₅₁

$ \Rightarrow t = {b_1}{{({2^{51}} - 1)} \over {2 - 1}}$ ..... (iii)

and s = a₁ + a₂ + .... + a₅₁

$ = {{51} \over 2}(2{a_1} + 50D)$ ...... (iv)

$\therefore$ t = a₁(2⁵¹ $-$ 1) [$\because$ a₁ = b₁]

or t = 2⁵¹ a₁ $-$ a₁ < 2⁵¹ a₁ ...... (v)

and $s = {{51} \over 2}[{a_1} + ({a_1} + 50D)]$ [from Eq. (ii)]

$ = {{51} \over 2}[{a_1} + {2^{50}}{a_1}] = {{51} \over 2}{a_1} + {{51} \over 2}{2^{50}}{a_1}$

$\therefore$ s > 2⁵¹ a₁ ...... (vi)

From Eqs. (v) and (vi),

we get s > t

Also, a₁₀₁ = a₁ + 100 D

and b₁₀₁ = 2¹⁰⁰ b₁

$\therefore$ ${a_{101}} = {a_1} + 100\left( {{{{2^{50}}{a_1} - {a_1}} \over {50}}} \right)$

and b₁₀₁ = 2¹⁰⁰ a₁

$\Rightarrow$ a₁₀₁ = a₁ + 2⁵¹ a₁ $-$ 2a₁ = 2⁵¹ a₁ $-$ a₁

$\Rightarrow$ a₁₀₁ < 2⁵¹ a₁

and b₁₀₁ > 2⁵¹ a₁ $\Rightarrow$ b₁₀₁ > a₁₀₁

Given: $a_1=5$ and $a_{20}=25$

Also given, $a_1, a_2, a_3, \ldots \ldots \ldots$ are in H.P.

$\Rightarrow \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \ldots \ldots$ are in A.P.

Let D be the common difference of above A. P.

$\begin{array}{ll} \therefore & \frac{1}{a_{20}}=\frac{1}{a_1}+(20-1) d \\ \Rightarrow & \frac{1}{25}=\frac{1}{5}+19 d \\ \Rightarrow & d=\frac{-4}{475} \end{array}$

$\begin{aligned} & \text { Now, } \quad \frac{1}{a_n}=\frac{1}{a_1}+(n-1) d \\ & \Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}+(n-1) \cdot\left(\frac{-4}{475}\right) \\ & \Rightarrow \quad \frac{1}{a_n}=\frac{95-4 n+4}{475} \\ & \Rightarrow \quad a_n=\frac{475}{99-4 n} \\ \end{aligned}$

Apply $\quad a_n<0$

$\Rightarrow \quad \frac{475}{99-4 n}<0$

The least positive integral value of $n$ is 25 which satisfy the above condition.

We have ${t_n} = c\{ {n^2} - {(n - 1)^2}\} $

$ = c(2n - 1)$

$ \Rightarrow t_n^2 = {c^2}(4{n^2} - 4n + 1)$

$ \Rightarrow \sum\limits_{n = 1}^n {t_n^2 = {c^2}\left\{ {{{4n(n + 1)(2n + 1)} \over 6} - {{4n(n + 1)} \over 2} + n} \right\}} $

$ = {{{c^2}n} \over 6}\{ 4(n + 1)(2n + 1) - 12(n + 1) + 6\} $

$ = {{{c^2}n} \over 3}\{ 4{n^2} + 6n + 2 - 6n - 6 + 3\} = {{{c^2}} \over 3}n(4{n^2} - 1)$

which is the sum of the square of $n$ terms.

Given, $a_1,a_2,a_3,a_4$ are in G.P.

Then, $b_1,b_2,b_3,b_4$ are the numbers.

$a_1,a_1+a_2,a_1+a_2+a_3,a_1+a_2+a_3+a_4$ or $a,a+ar,a+ar+ar^2,a+ar+ar^2+ar^3$

Clearly above numbers are neither in A.P. nor in G.P. and hence statement 1 is true.

Also, ${1 \over a},{1 \over {a + ar}},{1 \over {a + ar + a{r^2}}},{1 \over {a + ar + a{r^2} + a{r^3}}}$ are not in H.P.

$\therefore$ $b_1,b_2,b_3,b_4$ are not in H.P.

$\therefore$ Statement 2 is false.

If A, G, H be A.M, G.M and H.M then $\mathrm{G^{2}=A H}$ and $\mathrm{A} > \mathrm{G} > \mathrm{H}$

Also, we are given

$\mathrm{A}_{n-1}$ and $\mathrm{H}_{n-1}$ have A.M, G.M and H.M as $\mathrm{A}_{n}$, $\mathrm{G}_{n}$ and $\mathrm{H}_{n}$

$\therefore$ By definition, we have

$\mathrm{A}_{n}=\frac{\mathrm{A}_{n-1}+\mathrm{H}_{n-1}}{2}, \mathrm{G}_{n}^{2}=\mathrm{A}_{n-1} \text { and } \mathrm{H}_{n-1}$

$\frac{2}{\mathrm{H}_{n}}=\frac{1}{\mathrm{~A}_{n-1}}+\frac{1}{\mathrm{H}_{n-1}} \text { for }$

$n=2, \mathrm{G}_{2}^{2}=\mathrm{A}_{1} \mathrm{H}_{1}=\mathrm{G}_{1}^{2} \text { so on } $

$\therefore G_{1}^{2}=G_{2}^{2}=G_{3}^{2} \ldots \ldots$

$\Rightarrow \mathrm{G}_{1}=\mathrm{G}_{2}=\mathrm{G}_{3} \ldots \ldots$

$A_{2}$ is A.M. of $A_{1}, H_{1}$ and $A_{1} > H$

$\Rightarrow \mathrm{A}_{1} > \mathrm{A}_{2} > \mathrm{H}_{1}$

$\mathrm{A}_{3}$ is A.M. of $\mathrm{A}_{2}, \mathrm{H}_{2}$

$\mathrm{A}_{2} > \mathrm{A}_{3} > \mathrm{H}_{2}$

$\therefore \mathrm{A}_{1} > \mathrm{A}_{2} > \mathrm{A}_{3} \ldots \ldots$.

let $a$ and $b$ are two numbers then

$\begin{aligned}& \mathrm{A}_{1}=\frac{a+b}{2} ; \mathrm{G}_{1}=\sqrt{a b}, \mathrm{H}_{1}=\frac{2 a b}{a+b} \\ & \mathrm{~A}_{n}=\frac{\mathrm{A}_{n-1}+\mathrm{H}_{n-1}}{2} \end{aligned}$

$\mathrm{G}_{n}=\frac{2 \mathrm{~A}_{n-1}+\mathrm{H}_{n-1}}{\mathrm{~A}_{n-1}+\mathrm{H}_{n-1}}$

$G_{1}=G_{2}=G_{3}=\ldots . \cdot \sqrt{a b}$

$A_{2}$ is A.M. of $A_{1}$ and $H_{1}$ and $\mathrm{A}_{1} > \mathrm{H}_{1} \Rightarrow \mathrm{A}_{1} > \mathrm{A}_{2} > \mathrm{H}_{1}$

$A_{3}$ is A.M. of $A_{2}$ and $H_{2}$

$\mathrm{A}_{2} > \mathrm{H}_{2} \Rightarrow \mathrm{A}_{2} > \mathrm{A}_{3} > \mathrm{H}_{2}$ .......

$\therefore \mathrm{A}_{1} > \mathrm{A}_{2} > \mathrm{A}_{3} > \ldots$

$\mathrm{A}_{1} > \mathrm{H}_{2} > \mathrm{H}_{1} > \mathrm{A}_{2} > \mathrm{H}_{3} > \mathrm{H}_{2}$

$\therefore \mathrm{H}_{1} < \mathrm{H}_{2} < \mathrm{H}_{3} \ldots \ldots$.

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

$ = \sum\limits_{r = 1}^n {\left( {{r^3} - {{{r^2}} \over 2} + {r \over 2}} \right)} $

$ = \sum {{n^3} - {1 \over 2}\sum {{n^2} + {1 \over 2}\sum n } } $

$ = {{{n^2}{{(n + 1)}^2}} \over 4} - {{n(n + 1)(2n + 1)} \over {12}} + {{n(n + 1)} \over {2 \times 2}}$

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

$ = {{n(n + 1)[3{n^2} + n + 2]} \over {12}}$

$ = {1 \over {12}}n(n + 1)[3{n^2} + n + 2]$

We have,

${v_r} = {1 \over 2}(2{r^3} - {r^2} + r)$

${v_{r + 1}} = {1 \over 2}[2{(r + 1)^3} - {(r + 1)^2} + (r + 1)]$

Now,

${T_r} = {v_{r + 1}} - {v_r} - 2$

$ = [{(r + 1)^3} - {r^3}] - {1 \over 2}[{(r + 1)^2} - {r^2}] + {1 \over 2}(1) - 2$

$ = 3{r^2} + 2r - 1$

$ = (r + 1)(3r - 1)$

Which is a composite number

We have

${T_r} = 3{r^2} + 2r - 1$

${T_{r + 1}} = 3{(r + 1)^2} + 2(r + 1) - 1$

Now,

${Q_r} = {T_{r + 1}} - {T_r} = 3[{(r + 1)^2} - {r^2}] + 2(1)$

$ = 6r + 5$

${Q_{r + 1}} = 6(r + 1) + 5$

${Q_{r + 1}} - {Q_r} = 6 = $ constant

Therefore, Q$_1$, Q$_2$ and Q$_3$ ..... are in A.P. with common difference 6.

let $S_{n}$ be the sum of all.

run scored in $\mathrm{K}$ - matches.

i.e., $\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} k .2^{n+1-k}$

$\begin{aligned}& =2^{n+1} \sum_{k=1}^{n} k \cdot 2^{-k} \\ S_{n} & =2^{n+1}\left[\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\ldots . \cdot \frac{n}{2^{n}}\right]\quad .... \mathrm{(i)} \end{aligned}$

Multiply $S_{n}$ by $\frac{1}{2}$, we have.

$\frac{S_{n}}{2}=2^{n+1}\left[\frac{1}{2^{2}}+\frac{2}{2^{3}}+\frac{3}{2^{4}}+\ldots \ldots+\frac{n}{2^{n+1}}\right]$ ..... (ii)

Subtract (ii) from (i), we have.

$\begin{aligned} & \frac{S_{n}}{2}=2^{n+1}\left[\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots .+\frac{1}{2^{n}}-\frac{n}{2^{n+1}}\right] \\ &=2^{n+1}\left[\frac{\frac{1}{2}\left(1-\frac{1}{2^{n}}\right)}{\frac{1}{2}}-\frac{n}{2^{n+1}}\right] \\ &= 2^{n+1}\left[\frac{2^{n}-1}{2^{n}}-\frac{n}{2^{n+1}}\right] \\ &= {\left[2^{n+1}-2-n\right] } \\ & \Rightarrow \quad S_{n}= 2\left[2^{n+1}-2-n\right] \\ & \text { Now, } S_{n}= \frac{n+1}{4}\left(2^{n+1}-n-2\right) \text { given, } \\ & 2({2^{n + 1}} - 2 - n) = {{n + 1} \over 4}({2^{n + 1}} - n - 2) \\ & \Rightarrow \quad \frac{n+1}{4}=2\end{aligned}$

$\Rightarrow \quad n+1=8\quad \Rightarrow n=7$