Sequences and Series

As we know,

$ S_n=1^3+2^3+3^3+\ldots+n^3=\left[\frac{n(n+1)}{2}\right]^2 $

and $T_n=1+2+3+\ldots+n$

$ =\frac{n(n+1)}{2} $

Hence, $S_n=T_n^2$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } S_n=\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\frac{1}{7 \cdot 9}+\ldots \\ & S_{24}=\sum_{n=1}^{24} \frac{1}{(2 n+1)(2 n+3)} \\ & =\frac{1}{2} \sum_{n=1}^{24} \frac{1}{2 n+1}-\frac{1}{2 n+3} \\ & =\frac{1}{2}\left[\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)\right. \left.+\ldots+\left(\frac{1}{49}-\frac{1}{51}\right)\right] \\ & =\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\left(\frac{16}{51}\right)=\frac{8}{51} . \end{aligned} \end{aligned} $

$1+\frac{4}{15}+\frac{4 \cdot 10}{15 \cdot 30}+\frac{4 \cdot 10 \cdot 16}{15 \cdot 30 \cdot 45}+\ldots+\infty$

We know that

$ \begin{aligned} (1+x)^n=1+n x & +\frac{n(n-1)}{2!} x^2 \\ & +\frac{n(n-1)(n-2)}{3!} x^3+\ldots \end{aligned} $

On comparing with given expansion, we get

$ n x=\frac{4}{15} \Rightarrow x=\frac{4}{15 n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\frac{n(n-1)}{2!} x^2=\frac{4 \cdot 10}{15 \cdot 30}=\frac{40}{450}=\frac{4}{45}$

Using Eq. (i), $\frac{n(n-1)}{2!}\left(\frac{4}{15 n}\right)^2=\frac{4}{45}$

$ \begin{array}{lrl} \Rightarrow & \frac{n(n-1)}{2} \times \frac{4 \times 4}{15 n \times 15 n} & =\frac{4}{45} \\ \Rightarrow & \frac{n(n-1)}{2} \times \frac{16}{225 n^2} & =\frac{4}{45} \\ \Rightarrow & & n=\frac{-2}{3} \end{array} $

$ \begin{aligned} & \text { Also, } x=\frac{4}{15}\left(\frac{-3}{2}\right)=\frac{-2}{5} \\ & \begin{aligned} \therefore \quad(1+x)^n & =\left(1-\frac{2}{5}\right)^{-2 / 3} \\ & =\left(\frac{3}{5}\right)^{-2 / 3}=\left(\frac{5}{3}\right)^{2 / 3} \end{aligned} \end{aligned} $

Given, $t_n=\frac{1}{4}(n+2)(n+3), n \in N \,\,\,\,\,\,\,\,\,\,\,\,\,\ldots$ (i)

Also, $\frac{1}{t_1}+\frac{1}{t_2}+\ldots+\frac{1}{t_{2003}}=\frac{2003}{3009}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), we get

$ \begin{aligned} \frac{1}{t_n} & =\frac{4}{(n+2)(n+3)} \\ & =\frac{4}{n+2}-\frac{4}{n+3} \end{aligned} $

(By partial fraction)

$ \begin{aligned} \Rightarrow \quad & \sum_{n=1}^{2003} \frac{1}{t_n}=\sum_{n=1}^{2003}\left(\frac{4}{n+2}-\frac{4}{n+3}\right) \\ & =4\left(\frac{1}{3}-\frac{1}{2006}\right)=4\left(\frac{2006-3}{3 \times 2006}\right) \\ & =4 \times \frac{2003}{3 \times 2006} \\ & =\frac{8012}{6018}=\frac{4006}{3009} \end{aligned} $

So, Assertion is false.

Also, by putting $n=2003$ in Reason then we observe that it is wrong.

The integers divisible by 5 from 1 to 100 are $5,10,15,20,25, \ldots, 100$. Which are in A.P. with first term (a) $=5$ and common difference $(d)=5$ ) last term $(l)=100$

So, total number of terms,

$ \begin{aligned} (n) & =\frac{l-a}{d}+1=\frac{100-5}{5}+1=\frac{95}{5}+1 \\ & =19+1=20 \\ \therefore S_n & =\frac{n}{2}(a+l) \\ & =10(5+100)=1050 \end{aligned} $

Similarly, the integers divisible by 13 from 1 to 100 are $13,26,39,52, \ldots, 91$ which are in AP with to $a=13, d=13$, $l=91$.

$ \begin{aligned} & \text { } \begin{array}{l}So,\,\, n=\frac{l-a}{d}+1=\frac{91-13}{13}+1=\frac{78}{13}+1 \\ =6+1=7 \end{array} \\ & \therefore \quad S_n=\frac{7}{2}(13+91)=7 \times 52=364 \end{aligned} $

Now, the integer divisible by both 5 and 13 from 1 to 100 is 65 .

Hence, the sum of integer from 1 to 100 that are divisible by 5 or 13 is $=1050+364-65=1414-65=1349$

Given, $x>\sqrt{3} \Rightarrow x^2>3 \Rightarrow 0<\frac{1}{x^2}<\frac{1}{3}$ Also, $\frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{-1}{x^2+2}+\frac{2}{x^2+3} =(-1)\left(x^2+2\right)^{-1}+2\left(x^2+3\right)^{-1}$

$ \begin{aligned} &\begin{aligned} & =(-1)\left(x^2\right)^{-1}\left(1+\frac{2}{x^2}\right)^{-1}+2\left(x^2\right)^{-1}\left(1+\frac{3}{x^2}\right)^{-1} \\ & =\frac{-1}{x^2}\left(1-\frac{2}{x^2}+\frac{4}{x^4}-\frac{8}{x^6}+\frac{16}{x^8}-\ldots \infty\right) \\ & \quad+\frac{2}{x^2}\left(1-\frac{3}{x^2}+\frac{9}{x^4}-\frac{27}{x^6}+\ldots \infty\right) \end{aligned}\\ &\text { So, the coefficient of }\\ &\begin{aligned} x^{-8} & =(-1)(-8)+(2)(-27) \\ & =8-54=-46 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { From the given, we have }\\ &\begin{aligned} \sum_{k=1}^n k & (k+1)(k+2 \ldots(k+r-1) \\ & =\sum_{K=1}^n \frac{(k+r-1)!}{(k-1)!} \\ & =r!\sum_{k=1}^n{ }^{k+r-1} C_r \\ & =r!{ }^{n+r} C_{r+1} \\ & =\frac{n(n+1)(n+2) \ldots(n+r)}{(r+1)} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \mathrm{AM} \geq \mathrm{GM}\\ &\begin{array}{lc} \therefore & \frac{1+3+3^2+3^3+3^{n-1}}{n} \\ & \geq\left(1 \cdot 3 \cdot 3^2 \cdot 3^3 \cdot \ldots \cdot 3^{n-1}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2 n} \geq\left(3^{1+2+3+\ldots n-1}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2} \geq n\left(3^n \frac{(n-1)}{2}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2} \geq n 3^{\frac{n-1}{2}} \end{array} \end{aligned} $

$ \begin{aligned} &\text { Given, } 2 \cdot 5+5 \cdot 9+8 \cdot 13+11 \cdot 17+\ldots+\text { upto } n \text { terms }\\ &\begin{aligned} & t_n=(3 n-1)(4 n+1) \\ & t_n=12 n^2-n-1 \\ & \therefore S_n=\Sigma t_n=\Sigma\left(12 n^2-n-1\right)=12 \Sigma n^2-\Sigma n-\Sigma 1 \\ & =12 \cdot \frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}-n \\ & =n\left[2\left(2 n^2+3 n+1\right)-\frac{n+1}{2}-1\right]=\frac{n}{2}\left[8 n^2+12 n+4-n-1-2\right] \\ & =\frac{n}{2}\left[8 n^2+11 n+1\right] \\ & =4 n^3+\frac{11}{2} n^2+\frac{n}{2}=a n^3+b n^2+c n+d \\ & \Rightarrow a=4, b=\frac{11}{2}, c=\frac{1}{2} \text { and } d=0 \\ & \therefore a-b+c-d=4-\frac{11}{2}+\frac{1}{2}-0=4-5=-1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { We have, } 1^3+2^3+3^3+\ldots+n^3>x \\ & \because \quad \Sigma n^3=\left[\frac{n(n+1)}{2}\right]^2=\frac{n^2}{4}(n+1)^2>x \\ & \because \quad \frac{n^2}{4}<\frac{n^2}{4}(n+1)^2 \\ & \therefore \quad x=\frac{n^2}{4} \end{aligned} $

$\begin{array}{ll} \sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\ \sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)} \end{array}$

$\begin{aligned} & \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\ & \Rightarrow \quad \frac{(1-r)\left(1+r+r^2\right)}{(1-r)^3}=19 \\ & \Rightarrow \quad 18 r^2-39 r+18=0 \\ & \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\ & \therefore \quad a=19 \\ & \quad a+18 r \\ & \quad=19+12=31 \end{aligned}$

$\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5$

Multiply and divide by $d$

$\begin{aligned} & \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\ & \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots+\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\ & \frac{1}{d}\left[1-\frac{1}{1+10 d}\right]=5 \\ & \frac{1}{d}\left[\frac{1+10 d-1}{1+10 d}\right]=5 \end{aligned}$

$\begin{aligned} & \frac{10}{1+10 d}=5 \\ & 1+10 d=2 \\ & d=\frac{1}{10} \\ & 50 d=50 \times \frac{1}{10}=5 \end{aligned}$

Let's denote the first term of the geometric progression by $a$ and the common ratio by $r$. The terms of the geometric progression can be written as follows:

First term: $a$

Second term: $ar$

Third term: $(ar^2)$

Fourth term: $(ar^3)$

Fifth term: $(ar^4)$

Sixth term: $(ar^5)$

Eighth term: $(ar^7)$

We are given two key pieces of information:

1. The sum of the second and sixth terms is $\frac{70}{3}$:

$ar + ar^5 = \frac{70}{3}$

2. The product of the third and fifth terms is 49:

$(ar^2) \cdot (ar^4) = 49$

$a^2 r^6 = 49$

$a^2 = \frac{49}{r^6}$

$a = \frac{7}{r^3}$

Substituting $a = \frac{7}{r^3}$ into the first equation:

$\frac{7}{r^3} \cdot r + \frac{7}{r^3} \cdot r^5 = \frac{70}{3}$

$\frac{7r}{r^3} + \frac{7r^5}{r^3} = \frac{70}{3}$

$\frac{7}{r^2} + \frac{7r^2}{1} = \frac{70}{3}$

Let $x = r^2$. Then:

$\frac{7}{x} + 7x = \frac{70}{3}$

Multiply through by 3x to clear the denominator:

$21 + 21x^2 = 70x$

Rearrange into a standard quadratic equation:

$21x^2 - 70x + 21 = 0$

Divide by 7 to simplify:

$3x^2 - 10x + 3 = 0$

Solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = 3$, $b = -10$, and $c = 3$. Thus:

$x = \frac{10 \pm \sqrt{100 - 36}}{6}$

$x = \frac{10 \pm \sqrt{64}}{6}$

$x = \frac{10 \pm 8}{6}$

$x = 3$ or $x = \frac{1}{3}$

Since $x = r^2$ and $r$ is positive, we get $r = \sqrt{3}$ or $r = \frac{1}{\sqrt{3}}$. We need to choose the value that results in positive, increasing terms:

If $r = \sqrt{3}$:

$a = \frac{7}{r^3} = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}} = \frac{7}{3} \cdot \frac{1}{\sqrt{3}} = \frac{7\sqrt{3}}{9}$

Now we can determine the sum of the 4th, 6th, and 8th terms:

The 4th term is: $ar^3 = \frac{7\sqrt{3}}{9} \cdot 3\sqrt{3} = 7$

The 6th term is: $ar^5 = \frac{7\sqrt{3}}{9} \cdot 9\sqrt{3} = 21$

The 8th term is: $ar^7 = \frac{7\sqrt{3}}{9} \cdot 27(\sqrt{3}) = 49$

Adding these together:

$(4th + 6th + 8th terms) = 7 + 21 + 63 = 91$

Therefore, the sum of the 4th, 6th, and 8th terms is 91.

Correct answer:

Option C: 91

$\triangle A B C$ is an equilateral triangle having side $=a$ unit

Now, perimeter $=$ sum of all sides $=3 a$

Area $=\frac{\sqrt{3}}{4} a^2$

Now, $\ln \triangle D E F, D E=\frac{a}{2}=E F=D F$

$\begin{aligned} & \text { Perimeter }=3 \times \frac{a}{2}=\frac{3 a}{2} \\ & \text { Area }=\frac{\sqrt{3}}{4} \times\left(\frac{a}{2}\right)^2=\frac{\sqrt{3} a^2}{16} \end{aligned}$

Now, $P=3 a+\frac{3 a}{2}+\frac{3 a}{4}+\cdots$

$Q=\frac{\sqrt{3}}{4} a^2+\frac{\sqrt{3}}{16} a^2+\frac{\sqrt{3}}{64} a^2+\cdots$

$P=\frac{3 a}{1-\frac{1}{2}}=3 a \times 2 \Rightarrow P=6 a \quad \text{.... (i)}$

$Q=\frac{\frac{\sqrt{3}}{4} a^2}{1-\frac{1}{4}}=\frac{4}{3} \times \frac{\sqrt{3}}{4} a^2 \quad Q=\frac{\sqrt{3}}{3} a^2 \quad \text{.... (ii)}$

From equation (i) & (ii)

$\begin{aligned} & P=6 a \\ & Q=\frac{\sqrt{3}}{3} a^2 \\ & Q=\frac{\sqrt{3}}{3} \times \frac{P^2}{36} \\ & P^2=36 \sqrt{3} Q \end{aligned}$

To determine the value of $\mathrm{m}$, we need to formulate the problem using some basic concepts of arithmetic progression and work. Let's first understand the nature of the problem:

Initially, there are $\mathrm{m}$ computers, and it is estimated that with these $\mathrm{m}$ computers, the assignment can be completed in 17 days.

However, due to the crash of 4 computers every day starting from the second day onward, the total time taken extends by 8 days, making it 25 days in total.

To begin with, let's define the total work (W) in terms of the number of computers and days:

The total work (W) is given by:

The amount of work completed each day with $\mathrm{m}$ computers for 17 days:

$W = 17m$

When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are $\mathrm{m}$ computers. On the second day, there are $\mathrm{m} - 4$ computers, on the third day, there are $\mathrm{m} - 8$ computers, and so on. We need to sum this series until 25 days are completed.

This can be formulated as:

Total work done over 25 days with decrement in the number of computers:

$W = m + (m - 4) + (m - 8) + \ldots + \left[m - 4 \times (n - 1)\right]$

where $n$ is the number of days. Here, $n = 25$.

Notice that we form an arithmetic series where the first term (a) is $\mathrm{m}$ and the common difference (d) is -4. The sum of the first n terms of an arithmetic series is:

$S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]$

Plugging in the values:

$S_{25} = \frac{25}{2} \left[ 2m + (25 - 1)(-4) \right]$

$S_{25} = \frac{25}{2} \left[ 2m - 96 \right]$

$S_{25} = \frac{25}{2} \left[ 2m - 96 \right] = 25(m - 48)$

This work should be equivalent to the work calculated earlier, so:

$17m = 25(m - 48)$

Solving for $\mathrm{m}$:

$17m = 25m - 1200$

$8m = 1200$

$m = 150$

Thus, the value of $\mathrm{m}$ is equal to:

Option C: 150

To determine the least value of $\mathrm{K}$ for which the terms $4^{1+x} + 4^{1-x}, \frac{\mathrm{K}}{2}, 16^x + 16^{-x}$ form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.

For three numbers to be in an arithmetic progression, the middle term must be the average of the other two terms. Therefore, we can write:

$ \frac{4^{1+x} + 4^{1-x} + 16^x + 16^{-x}}{2} = \frac{K}{2} $

First, simplify each term individually:

1. Consider $4^{1+x} + 4^{1-x}$:

$4^{1+x} = 4 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x} = 4 \cdot 2^{2x}$

and

$4^{1-x} = 4 \cdot 4^{-x} = 4 \cdot (2^2)^{-x} = 4 \cdot 2^{-2x} = 4 \cdot 2^{-2x}$

Thus,

$4^{1+x} + 4^{1-x} = 4 \cdot 2^{2x} + 4 \cdot 2^{-2x} = 4(2^{2x} + 2^{-2x})$

2. Consider $16^x + 16^{-x}$:

$16^x = (2^4)^x = 2^{4x}$

and

$16^{-x} = (2^4)^{-x} = 2^{-4x}$

Thus,

$16^x + 16^{-x} = 2^{4x} + 2^{-4x}$

3. Combine the terms and set up the equation:

$ \frac{4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x}}{2} = \frac{K}{2} $

Multiply both sides by 2:

$ 4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x} = K $

To find the least value of $\mathrm{K}$, let's assume $x = 0$ (since $x$ can range over non-negative values):

For $x = 0$:

$4(2^{2 \cdot 0} + 2^{-2 \cdot 0}) + 2^{4 \cdot 0} + 2^{-4 \cdot 0}$

This simplifies to:

$4(2^0 + 2^0) + 2^0 + 2^0$

$ = 4(1 + 1) + 1 + 1 $

$ = 4 \cdot 2 + 1 + 1 $

$ = 8 + 1 + 1 $

$ = 10 $

Therefore, the least value of $\mathrm{K}$ that ensures the values form an arithmetic progression is $ 10 $. Hence, the correct option is:

Option A: 10

$\begin{aligned} & \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m \\ & \text { and } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n \\ & \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \\ & =\frac{1}{\sqrt{1}+\sqrt{2}} \times \frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ & \quad+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \times \frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}} \end{aligned}$

$\begin{aligned} & =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{100}-\sqrt{99} \\ & =\sqrt{100}-\sqrt{1} \\ & =10-1 \\ & \Rightarrow m=9 \end{aligned}$

and $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$

$\begin{aligned} & \frac{2-1}{1 \times 2}+\frac{3-2}{2 \times 3}+\ldots+\frac{100-99}{100 \times 99}=n \\ & \Rightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots+\frac{1}{99}-\frac{1}{100}=n \\ & \Rightarrow n=1-\frac{1}{100} \\ & \Rightarrow n=\frac{99}{100} \\ & (m, n)=\left(9, \frac{99}{100}\right) \end{aligned}$

Satisfies the line $11 x-100 y=0$

$\begin{aligned} & \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101} \\ & \Rightarrow \frac{\sum_\limits{n=1}^{100} n(n+1)^2}{\sum_\limits{n=1}^{100} n^2(n+1)} \end{aligned}$

$\begin{aligned} & \Rightarrow \frac{\sum_\limits{n=1}^{100} n^3+2 n^2+n}{\sum_\limits{n=1}^{100} n^3+n^2} \\ & =\frac{\left(\frac{100(101)}{2}\right)^2+\frac{2 \cdot 100(101)(201)}{6}+\frac{100(101)}{2}}{\left(\frac{100(101)}{2}\right)^2+\frac{100(101)(201)}{6}} \\ & =\frac{300(101)+4(201)+6}{300(101)+2(201)}=\frac{5185}{5117}=\frac{305}{301} \end{aligned}$

$\begin{aligned} & 2 b=a+c \quad \text{.... (1)}\\ & b^2=(a+1)(c+3) \quad \text{.... (2)}\\ & \frac{a+b+c}{3}=8 \quad \text{.... (3)} \end{aligned}$

$\begin{aligned} \Rightarrow & \frac{3 b}{3}=8 \\ & b=8 \\ \Rightarrow \quad & a c+3 a+c+3=64 \end{aligned}$

$\begin{aligned} & 3 a+c+a c=61 \quad \text{... (4)}\\ & a+c=16 \\ & c=16-a \end{aligned}$

from equation (4)

$\begin{aligned} & 3 a+16-a+a(16-a)=61 \\ & \Rightarrow \quad(a-15)(a-3)=0 \\ & \quad a=15(a>10) \\ & \Rightarrow \quad a=15, b=8, c=1 \\ & \left((a \cdot b \cdot c)^{\frac{1}{3}}\right)^3=15 \times 8 \times 1=120 \end{aligned}$

$\begin{aligned} & \text { Let } p=2 r, q=2 r^2 \\ & T_7=2, T_8=2 r, T_{13}=2 r^2 \\ & d=2 r-2=2(r-1) \\ & 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \\ & \Rightarrow r^2-6 r+5=0 \\ & \Rightarrow(r-1)(r-5)=0 \\ & \therefore r=1,5 \\ & r=1 \text { (rejected) as } q \neq 2 \\ & \therefore r=5 \end{aligned}$

$5^{\text {th }}$ term of G.P $=2 . r^4=2.5^4$

Let $1^{\text {st }}$ term of A.P $b a=a, d=8$

$2=a+(6)(8) \Rightarrow a=-46$

$\mathrm{n}^{\text {th }}$ term of A.P $=-46+(n-1) 8=8 n-54$

$\begin{aligned} & 2.5^4=8 n-54 \\ & \Rightarrow 1250+54=8 n \\ & \Rightarrow n=\frac{1304}{8}=163 \end{aligned}$

To solve this problem, we will start by using the properties of an arithmetic progression (AP).

The sum of the first $n$ terms of an AP can be calculated using the formula: $ S_n = \frac{n}{2} (2a + (n-1)d) $ where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference between the terms.

Given the information: $ S_{10} = 390 $

We can plug $n=10$ into the sum formula to get:

$ S_{10} = \frac{10}{2} (2a + (10-1)d) $

$ 390 = 5(2a + 9d) $

$ 390 = 10a + 45d $

$ 78 = 2a + 9d \quad .........\text{(1)} $

Next, we're given the ratio of the tenth term ($T_{10}$) to the fifth term ($T_5$): $ \frac{T_{10}}{T_5} = \frac{15}{7} $

The $n$th term of an AP is given by:

$ T_n = a + (n-1)d $

So, for the tenth term: $ T_{10} = a + (10-1)d = a + 9d $

And for the fifth term:

$ T_5 = a + (5-1)d = a + 4d $

Now we can write the ratio as:

$ \frac{a + 9d}{a + 4d} = \frac{15}{7} $

$ 7(a + 9d) = 15(a + 4d) $

$ 7a + 63d = 15a + 60d $

$ 63d - 60d = 15a - 7a $

$ 3d = 8a \quad .........\text{(2)} $

Now we have two equations (1) and (2):

$ 78 = 2a + 9d \quad \text{(1)} $

$ 3d = 8a \quad \text{(2)} $

We can solve these equations simultaneously.

From equation (2):

$ d = \frac{8}{3}a $

Plugging this back into (1):

$ 78 = 2a + 9\left(\frac{8}{3}a\right) $

$ 78 = 2a + 24a $

$ 78 = 26a $

$ a = 3 $

Now we can find $d$:

$ d = \frac{8}{3}a $

$ d = \frac{8}{3} \times 3 $

$ d = 8 $

Now we can find $S_{15}$ and $S_5$ using the formula for the sum of an AP.

For $S_{15}$:

$ S_{15} = \frac{15}{2} (2 \cdot 3 + (15-1) \cdot 8) $

$ S_{15} = \frac{15}{2} (6 + 14 \cdot 8) $

$ S_{15} = \frac{15}{2} (6 + 112) $

$ S_{15} = \frac{15}{2} \cdot 118 $

$ S_{15} = 15 \cdot 59 $

$ S_{15} = 885 $

For $S_5$:

$ S_5 = \frac{5}{2} (2 \cdot 3 + (5-1) \cdot 8) $

$ S_5 = \frac{5}{2} (6 + 4 \cdot 8) $

$ S_5 = \frac{5}{2} (6 + 32) $

$ S_5 = \frac{5}{2} \cdot 38 $

$ S_5 = 5 \cdot 19 $

$ S_5 = 95 $

The difference $S_{15} - S_{5}$ is:

$ S_{15} - S_{5} = 885 - 95 $

$ S_{15} - S_{5} = 790 $

Therefore, the correct answer is Option C, which is 790.

Since $3, a, b, c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:

$ d = a - 3 $

The nth term of an A.P. is given by the formula:

$ T_n = a + (n-1)d $

So, using this formula, we can express $b$ and $c$ in terms of $a$ and $d$:

$ b = a + d $

$ c = a + 2d $

Substituting $d = a - 3$ into these expressions:

$ b = a + (a - 3) $

$ c = a + 2(a - 3) $

Therefore:

$ b = 2a - 3 $

$ c = 3a - 6 $

Now, let's consider that $3, a-1, b+1, c+9$ are in geometric progression (G.P.). For terms in a G.P., the ratio (common ratio, r) between consecutive terms is constant. So:

$ \frac{a - 1}{3} = \frac{b + 1}{a - 1} = \frac{c + 9}{b + 1} $

Now, we will establish the relation between the terms using the property of G.P.:

$ \frac{a - 1}{3} = \frac{b + 1}{a - 1} $

$ (a - 1)^2 = 3(b + 1) $

$ a^2 - 2a + 1 = 3b + 3 $

Substituting $b = 2a - 3$, we get:

$ a^2 - 2a + 1 = 3(2a - 3) + 3 $

$ a^2 - 2a + 1 = 6a - 9 + 3 $

$ a^2 - 8a + 7 = 0 $

Solving this quadratic equation:

$ (a - 7)(a - 1) = 0 $

Hence, $a = 7$ or $a = 1$. However, if $a = 1$, the terms $3, a-1, b+1, c+9$ cannot form a G.P. as it would involve division by zero. Therefore, $a = 7$. We use this value to find $b$ and $c$:

$ b = 2a - 3 = 2(7) - 3 = 14 - 3 = 11 $

$ c = 3a - 6 = 3(7) - 6 = 21 - 6 = 15 $

Now we can find the arithmetic mean ($A$) of $a$, $b$, and $c$:

$ A = \frac{a + b + c}{3} $

$ A = \frac{7 + 11 + 15}{3} $

$ A = \frac{33}{3} $

$ A = 11 $

Hence, the arithmetic mean of $a$, $b$, and $c$ is $11$, which corresponds to Option D.

$\begin{aligned} & 1+d, \quad 1+7 d, 1+43 d \text { are in GP } \\ & (1+7 d)^2=(1+d)(1+43 d) \\ & 1+49 d^2+14 d=1+44 d+43 d^2 \\ & 6 d^2-30 d=0 \\ & d=5 \\ & S_{20}=\frac{20}{2}[2 \times 1+(20-1) \times 5] \\ & \quad=10[2+95] \\ & \quad=970 \end{aligned}$

$\begin{aligned} & f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\ & f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\ & f(1)=0 \\ & \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\ & \alpha=\frac{c+a-2 b}{a+b-2 c} \\ & \text { If, }-1<\alpha<0 \\ & -1<\frac{c+a-2 b}{a+b-2 c}<0 \\ & b+c<2 a \text { and } b>\frac{a+c}{2} \end{aligned}$

therefore, b cannot be G.M. between a and c.

$\begin{aligned} & \text { If, } 0<\alpha<1 \\ & 0<\frac{c+a-2 b}{a+b-2 c}<1 \\ & b>c \text { and } b<\frac{a+c}{2} \end{aligned}$

Therefore, $\mathrm{b}$ may be the G.M. between $\mathrm{a}$ and $\mathrm{c}$.

General term of the sequence,

$\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-1\right)^2-\mathrm{r}^2} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\ & \mathrm{T}_{\mathrm{r}}=\frac{\frac{1}{2}\left[\left(\mathrm{r}^2+\mathrm{r}-1\right)-\left(\mathrm{r}^2-\mathrm{r}-1\right)\right]}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\ & =\frac{1}{2}\left[\frac{1}{\mathrm{r}^2-\mathrm{r}-1}-\frac{1}{\mathrm{r}^2+\mathrm{r}-1}\right] \end{aligned}$

Sum of 10 terms,

$\sum_\limits{\mathrm{r}=1}^{10} \mathrm{~T}_{\mathrm{r}}=\frac{1}{2}\left[\frac{1}{-1}-\frac{1}{109}\right]=\frac{-55}{109}$

The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value. We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified to be equal.

For the first GP, with first term $a$ and third term $b$:

• The third term is given by $t_3 = ar^2 = b$, leading to $r^2 = \frac{b}{a}$.
• The eleventh term is $t_{11} = ar^{10} = a\left(\frac{b}{a}\right)^5$.

For the second GP, with first term $a$ and fifth term $b$:

• The fifth term is $T_5 = ar^4 = b$, yielding $r^4 = \frac{b}{a}$ and $r = \left(\frac{b}{a}\right)^{1/4}$.
• The $p^{\text{th}}$ term is thus $T_p = ar^{p-1} = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$.

Equating the eleventh term of the first GP to the $p^{\text{th}}$ term of the second GP gives: $a\left(\frac{b}{a}\right)^5 = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$.

Simplifying, we find that $5 = \frac{p-1}{4}$, leading to $p = 21$, which is the solution.

$\begin{aligned} &\begin{aligned} & \mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790 \\ & 2 \mathrm{a}+19 \mathrm{~d}=79 \quad \text{.... (1)}\\ & \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 \\ & 2 \mathrm{a}+9 \mathrm{~d}=29 \quad \text{.... (2)} \end{aligned}\\ &\text { From (1) and (2) } a=-8, d=5 \end{aligned}$

$\begin{aligned} & S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\ & =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\ & =405-10 \\ & =395 \end{aligned}$

$\log _{\mathrm{e}} \mathrm{a}, \log _{\mathrm{e}} \mathrm{b}, \log _{\mathrm{e}} \mathrm{c}$ are in A.P.

$\therefore \mathrm{b}^2=\mathrm{ac}$ ..... (i)

Also

$\begin{aligned} & \log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right) \text { are in A.P. } \\ & \left(\frac{2 b}{3 c}\right)^2=\frac{a}{2 b} \times \frac{3 c}{a} \\ & \frac{b}{c}=\frac{3}{2} \end{aligned}$

Putting in eq. (i) $b^2=a \times \frac{2 b}{3}$

$\begin{aligned} & \frac{\mathrm{a}}{\mathrm{b}}=\frac{3}{2} \\ & \mathrm{a}: \mathrm{b}: \mathrm{c}=9: 6: 4 \end{aligned}$

Let $r^{\prime}$th term of the GP be $a^{n-1}$. Given,

$\begin{aligned} & 2 a_r=a_{r+1}+a_{r+2} \\ & 2 a r^{n-1}=a r^n+a r^{n+1} \\ & \frac{2}{r}=1+r \\ & r^2+r-2=0 \end{aligned}$

Hence, we get, $r=-2$ (as $r \neq 1$)

So, $\mathrm{S}_{20}-\mathrm{S}_{18}=$ (Sum upto 20 terms) $-$ (Sum upto 18 terms) $=\mathrm{T}_{19}+\mathrm{T}_{20}$

$\mathrm{~T}_{19}+\mathrm{T}_{20}=\mathrm{ar}^{18}(1+\mathrm{r})$

Putting the values $\mathrm{a}=\frac{1}{8}$ and $\mathrm{r}=-2$;

we get $T_{19}+T_{20}=-2^{15}$

$\begin{aligned} & a+a r+a r^2+a r^3+\ldots .+a r^{63} \\ & =7\left(a+a r^2+a r^4 \ldots .+a r^{62}\right) \\ & \Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2} \\ & r=6 \end{aligned}$

$\begin{aligned} & a_6=2 \Rightarrow a+5 d=2 \\ & a_1 a_4 a_5=a(a+3 d)(a+4 d) \\ & =(2-5 d)(2-2 d)(2-d) \\ & f(d)=8-32 d+34 d^2-20 d+30 d^2-10 d^3 \\ & f^{\prime}(d)=-2(5 d-8)(3 d-2) \end{aligned}$

$\mathrm{d}=\frac{8}{5}$

$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}$

This is A.P. with common difference

$\begin{aligned} & d_1=-1+\frac{1}{4}=-\frac{3}{4} \\ & -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20 \end{aligned}$

This is also A.P. $\mathrm{a}=-129 \frac{1}{4}$ and $\mathrm{d}=\frac{3}{4}$

Required term $=$

$\begin{aligned} & -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\ & =-129-\frac{1}{4}+15-\frac{3}{4}=-115 \end{aligned}$

$4,9,14,19, \ldots$, up to $25^{\text {th }}$ term

$\mathrm{T}_{25}=4+(25-1) 5=4+120=124$

$3,6,9,12, \ldots$, up to $37^{\text {th }}$ term

$\mathrm{T}_{37}=3+(37-1) 3=3+108=111$

Common difference of $\mathrm{I}^{\text {st }}$ series $\mathrm{d}_1=5$

Common difference of $\mathrm{II}^{\text {nd }}$ series $\mathrm{d}_2=3$

First common term $=9$, and their common difference $=15\left(\operatorname{LCM}\right.$ of $\mathrm{d}_1$ and $\left.\mathrm{d}_2\right)$ then common terms are $9,24,39,54,69,84,99$

$ \begin{aligned} & \text t_n=(2 n-1)(2 n+1)(2 n+3) \\\\ & t_n=8 n^3+12 n^2-2 n-3 \\\\ & S_n=\Sigma S_n=8 \Sigma n^3+12 \Sigma n^2-2 \Sigma n-\Sigma 3 \\\\ & =8\left(\frac{n(n+1)}{2}\right)^2+12 \frac{n(n+1)(2 n+1)}{6} -2 \frac{n(n+1)}{2}-3 n \\\\ & =n\left[\frac{8 \cdot n(n+1)^2}{4}+2(n+1)(2 n+1)-n-1-3\right] \\\\ & =n\left[2 n\left(n^2+1+2 n\right)+2\left(2 n^2+3 n+1\right)-n-4\right] \\\\ & =n\left[2 n^3+4 n^2+2 n+4 n^2+6 n+2-n-4\right] \\\\ & =n\left[2 n^3+8 n^2+7 n-2\right] \\\\ & =n(n+1) f(n) \\\\ & \therefore f(n)=\frac{2 n^3+8 n^2+7 n-2}{n+1} \\\\ & \quad f(2)=\frac{16+32+14-2}{3}=\frac{60}{3}=20 \end{aligned} $

The given series is related to the binomial expansion for any index. The binomial expansion for $(1-x)^{-n}$ when $|x| < 1$ is given by :

$ (1-x)^{-n} = 1 + nx + \frac{n(n+1)}{1 \cdot 2}x^2 + \frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3}x^3 + \ldots $

To relate the series with the binomial expansion, consider setting $x = \frac{1}{2}$ and $n = \frac{2}{3}$. The substitution into the binomial formula results in :

$ \left(\frac{1}{2}\right)^{-\frac{2}{3}} = 1 + \frac{2 \cdot 1}{3 \cdot 2} + \frac{2 \cdot 5 \cdot 1}{3 \cdot 6 \cdot 4} + \ldots $

This series equates to $\sqrt[3]{4}$ :

$ \sqrt[3]{4} = 1 + \frac{2 \cdot 1}{3 \cdot 2} + \frac{2 \cdot 5 \cdot 1}{3 \cdot 6 \cdot 4} + \ldots $

Thus, the assertion that this infinite series sums to $\sqrt[3]{4}$ is correct, and it is derived from the application of the binomial series expansion with the appropriate substitutions for $x$ and $n$.

We have to check each option, From option (a), we have,

$ \begin{array}{l} (2 n+7) < (n+3)^{2} \\\\ \text { Let } P(n) \equiv(2 n+7) < (n+3)^{2} \\\\ \qquad P(1)=(2+7)=9 < (1+3)^{2}=4^{2}=16 \end{array} $

Let us assume that $ P(K) $ is true.

We have to prove that $ P(K+1) $ is true

$ \begin{array}{l} (2 K+7) < (K+3)^{2} \\\\ 2 K+7 < K^{2}+9+6 K \end{array} $

Add 2 on LHS and $ 7+2 K $ on RHS

as $ 2 < 7+2 K $

$ \begin{array}{l} \Rightarrow 2 K+9 < K^{2}+16+8 K \\\\ 2(K+1)+7 < ((K+1)+3)^{2} \end{array} $

$ \therefore P(K+1) $ is true.

Hence, $ P(x) $ is true, $ \forall n \in N $.

From option b,

$ \begin{array}{r} \text { Let } Q(n) \equiv 1^{2}+2^{2}+\ldots+n^{2} > \frac{n^{3}}{3} \\\\ \qquad Q(1)=1^{2} > \frac{1}{3} \end{array} $

Let $ Q(K) $ be true.

$ \begin{array}{c} 1^{2}+2^{2}+3^{2}+\ldots+K^{2} > \frac{K^{3}}{3} \\\\ 1^{2}+2^{2}+3^{2}+\ldots .+K^{2} \\\\ +(K+1)^{2} > \frac{K^{3}}{3}+(K+1)^{2} \\\\ 1^{2}+2^{2}+3^{2}+\ldots .+K^{2}+(K+1)^{2} \\\\ > \frac{K^{3}+3 K^{2}+6 K+3}{3} \\\\ 1^{2}+2^{2}+3^{2}+\ldots+K^{2}+(K+1)^{2} > \\\\ \frac{K^{3}+3 K^{2}+3 K+1+3 K+2}{3} \\\\ =\frac{(K+1)^{3}+3 K+2}{3} > \frac{(K+1)^{3}}{3} \end{array} $

$ \therefore Q(K+1)^{2} $ is true.

Hence, $ Q(n) $ is true, $ \forall n \in N $

From option (c),

$ \begin{array}{l} R(n)=35^{2 n+1}+2^{3 n+1} \\\\ R(1)=3 \times 125+2^{4}=375+16=391 \end{array} $

which is divisible by 23

Let us assume it is true for $ n=k, k \in N $

$ \therefore S(k)=3 \cdot 5^{2 k+1}+2^{3 k+1} $ is divisible by

$ 23 \forall k \in N $

Let $ 3 \cdot 5^{2 k+1}+2^{3 k+1}=23 t, t \in N $

$ \therefore 3 \cdot 5 k^{2 k+1}=23 t-2^{3 k+1} $

Now, we have to prove that for $ n=k+1 $

$ \begin{aligned} \therefore S(k+1) & =3 \cdot 5^{2(k+1)+1}+2^{3(k+1)+1} \\\\ & =3 \cdot 5^{2 k+1} \cdot 5^{2}+2^{3 k+1} \cdot 2^{3} \\\\ & =\left(23 t-2^{3 k+1}\right) 25+8 \cdot 2^{3 k+1} \\\\ & =23 t \times 25-25 \times 2^{3 k+1}+8 \cdot 2^{3 k+1} \\\\ \therefore S(k+1) & =23 t \times 25-172^{3 k+1} \end{aligned} $

Hence, $ 3 \cdot 5^{2 n+1}+2^{3 n+1} $ is not divisible by 23.

Hence, $ R(n) $ is not divisible by $ 23, \forall n \in N $. From option (d),

$ 2+7+12+\ldots+(5 n-3) $ is an AP with common difference as 5 .

$ \begin{aligned} S_{n} & =\frac{n}{2}[2 \times 2+(n-1) 5] \\\\ & =\frac{n}{2}(5 n-1) \end{aligned} $

$\begin{aligned} \frac{1}{3 \cdot 6} & +\frac{1}{6 \cdot 9}+\frac{1}{9 \cdot 12}+\ldots \text { to } 9 \text { terms } \\ = & \frac{1}{9}\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{9 \cdot 10}\right) \\ = & \frac{1}{9}\left(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)\right. \left.+\ldots+\left(\frac{1}{9}-\frac{1}{10}\right)\right) \\ & = \left(1-\frac{1}{10}\right)=\frac{1}{9}\left(\frac{9}{10}\right)=\frac{1}{10}\end{aligned}$