Sequences and Series

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.

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

$\begin{aligned} & \text { Given, } \mathrm{S}_n=\sum_{k=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot \mathrm{K}^2 \\ & \Rightarrow \mathrm{S}_n=-1^2-2^2+3^2+4^2-5^2-6^2+7^2 +8^2-9^2-10^2+11^2+12^2 \ldots \ldots . 4 n \text { terms } \end{aligned}$

$\begin{aligned} \Rightarrow \mathrm{S}_n= & -\left[1^2+5^2+9^2+\ldots . n \text { terms }\right] \\ & -\left[2^2+6^2+10^2+\ldots n \text { terms }\right] \\ & +\left[3^2+7^2+11^2+\ldots n \text { terms }\right] \\ & +\left[4^2+8^2+12^2+\ldots n \text { terms }\right] \end{aligned}$

$\Rightarrow \mathrm{S}_n=-\sum_\limits{r=1}^n(4 r-3)^2-\sum_\limits{r=1}^n(4 r-2)^2 +\sum_\limits{r=1}^n(4 r-1)^2+\sum_\limits{r=1}^n(4 r)^2$

$\begin{aligned} & \Rightarrow \mathrm{S}_n=\sum_{r=1}^n\left((4 r)^2+(4 r-1)^2-(4 r-2)^2-(4 r-3)^2\right) \\ & \Rightarrow \mathrm{S}_n=\sum_{r=1}^n(32 r-12) \\ & \Rightarrow \mathrm{S}_n=32 \sum_{r=1}^n r-\sum_{r=1}^n 12 \\ & \Rightarrow \mathrm{S}_n=32 \cdot \frac{n(n+1)}{2}-12 n \\ & \Rightarrow \mathrm{S}_n=16 n^2+16 n-12 n \\ & \Rightarrow \mathrm{S}_n=4 n(4 n+1) \end{aligned}$

If $n=9$, then $\mathrm{S}_9=1332$

If $n=8$, then $\mathrm{S}_8=1056$

Hints :

(i) Recall $\sum_\limits{\mathrm{K}=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot \mathrm{K}^2=-\sum_\limits{r=1}^n(4 r-3)^2-\sum_\limits{r=1}^n(4 r-2)^2+\sum_\limits{r=1}^n(4 r-1)^2+\sum_\limits{r=1}^n(4 r)^2$

(ii) $\sum_\limits{r=1}^n a \cdot f(r)+b \cdot g(r)-c \cdot h(r) = \sum_\limits{r=1}^n a \cdot f(r)+\sum_\limits{r=1}^n b \cdot g(r)-\sum_\limits{r=1}^n c \cdot h(r)$

$=a \sum_\limits{r=1}^n f(r)+b \sum_\limits{r=1}^n g(r)-c \sum_\limits{r=1}^n h(r)$

(iii) $\sum_\limits{r=1}^n r=\frac{n(n+1)}{2}, \sum_\limits{r=1}^n a=a r$ where is a constant.

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.

${S_n} < \mathop {\lim }\limits_{x \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{1 \over n}{1 \over {1 + {k \over n} + \left( {{k \over {{n^2}}}} \right)}}} $

$ = \int\limits_0^1 {{{dx} \over {1 + x + {x^2}}} = {\pi \over {3\sqrt 3 }}} $

As

$h\sum\limits_{k = 0}^n {f(kh) > \int\limits_0^1 {f(x)dx > h} } $

$\sum\limits_{k = 1}^n {f(kh)} $

So, ${T_n} > {\pi \over {3\sqrt 3 }}$

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 $H_{2}$

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

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

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

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