Using the arithmetic progression formula for the nth term, where

$x_n=a+\left(n-1\right)d$

Substituting the value for n and using that in the equation that is given, 

$2x_{6}+2x_{9}=x_{11}+x_{13}$, Then,$x_{100}$ equals

We get, $2\left(a+5d\right)+2\left(a+8d\right)=a+10d+a+12d$

$4a+26d=2a+22d$

$2a=-4d$

$a=-2d$

We are given, $x_5=-4$

$a+4d=-4$

Substituting the value for a in terms of d, 

$2d=-4$

$d=-2$

$a=4$

$x_{100}=a+99d$

$x_{100}=4-198=-194$

We are told that, for any natural number $n_{1}$ let $a_{n}$ be the largest integer not exceeding $\sqrt{n}$

So for n=1, the largest integer not exceeding $\sqrt{1}$ will be 1

For n=2, the largest integer not exceeding $\sqrt{2}$ will be 1

For n=3, the largest integer not exceeding $\sqrt{3}$ will be 1

For n=4, the largest integer not exceeding $\sqrt{4}$ will be 2

We see a pattern here regarding the squares of the numbers, 

Listing down all the perfect squares, 

1, 4, 9, 16, 25, 36, 49, 64, ...

We see that the difference between 4 and 1 is 3 and there were three natural numbers in the given pattern with the value as 1, 

So we can write for the rest of the numbers as well, 

3 numbers will have value 1, giving a total value of 3

5 numbers will have value 2, giving a total value of 10

7 numbers will have value 3, giving a total value of 21

9 numbers will have value 4, giving a total value of 36

11 numbers will have value 5, giving a total value of 55

13 numbers will have value 6, giving a total value of 78

Now, only the values of $a_{49},\ a_{50}$ will have the value of 7, total value of 14. 

Adding these values, we get the total sum as 217, which is the answer. 

Opening the brackets, we get the series as: $\left(\dfrac{1}{5}\right)^2-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)+\left(\dfrac{1}{5}\right)^4-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2+\left(\dfrac{1}{5}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^3+...$

These are two infinite GPs when rearranged:

$\left(\dfrac{1}{5}\right)^2+\left(\dfrac{1}{5}\right)^4+\left(\dfrac{1}{5}\right)^6+...-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^3-...$

The sum of the first series would be $\dfrac{\dfrac{1}{25}}{1-\dfrac{1}{25}}=\dfrac{1}{24}$

The sum of the second series would be $\frac{\dfrac{1}{35}}{1-\dfrac{1}{35}}=\dfrac{1}{34}$

The answer to the given series would then be $\dfrac{1}{24}-\dfrac{1}{34}=\dfrac{10}{816}=\dfrac{5}{408}$

Therefore, Option B is the correct answer. 

Finding the terms in the sequence, we see that $t_3=0$, $t_4=-\dfrac{1}{3}$, $t_5=0$

We would notice that all the odd terms are 0, and we are also asked the sum of only even terms, so we do not need to consider those

$t_6=-\dfrac{1}{5}$

We see that the even terms are in an HP: $-1,\ -\dfrac{1}{3},\ -\dfrac{1}{5},\ -\dfrac{1}{7},\ ...$

The sum we are asked is the inverse of these terms, that is: -1, -3, -5, -7, up to 1012 terms

The sum of this AP would be $\dfrac{\left[-\left(2\times\ 1\right)+\left(1012-1\right)\left(-2\right)\right]}{2}\times\ 1012$

Which is equal to $-1012\times\ 1012\ =\ -1024144$

Therefore, Option A is the correct answer. 

Given that $ \frac{1}{\sqrt{y} + \sqrt{z}} $ is the arithmetic mean of $ \frac{1}{\sqrt{x} + \sqrt{z}} $ and $ \frac{1}{\sqrt{x} + \sqrt{y}} $

=> $ \frac{2}{\sqrt{y} + \sqrt{z}} = \frac{1}{\sqrt{x} + \sqrt{z}} + \frac{1}{\sqrt{x} + \sqrt{y}} $

=> $ 2 \left(\sqrt{x} + \sqrt{z}\right) \left(\sqrt{x} + \sqrt{y}\right) = \left(\sqrt{y} + \sqrt{z}\right) \left(\sqrt{x} + \sqrt{y} + \sqrt{x} + \sqrt{z}\right) $

=> $ 2 \left(x + \sqrt{xy} + \sqrt{xz} + \sqrt{yz}\right) = 2\sqrt{xy} + y + \sqrt{yz} + 2\sqrt{xz} + \sqrt{yz} + z $

=> $ 2x = y + z $

=> y, x, z are in A.P. as x is the arithmetic mean of y and z.

Given on day-1, there are 2 organisms.

On day-2, there are 2*2 + 3 = 7 and on day-3, there are 2*7 + 3 = 17...

Let us try to form a pattern:

2 = 2 + 0 (n = 1)

7 = 4 + 3 (n = 2)

17 = 8 + 9 [8 + 3*3] (n = 3)

37 = 16 + 21[16 + 3*7] n = 4

T(n) = $ 2^n + 3 \left(2^{n-1} - 1\right) $

We know that $ 2^{20} = 2^{10} \times 2^{10} = 1024 \times 1024 $, which is more than 1 million.

Let us check for n = 19

$ 2^{19} + 3 \left(2^{18} - 1\right) = 2^{19} + 3 \cdot 2^{18} - 3 = 2 \cdot 2^{19} + 2^{18} - 3 = 2^{20} + 2^{18} - 3 $, which is more than 1 million.

Let us check for n = 18

=> $ 2^{18} + 3 \left(2^{17} - 1\right) = 2^{18} + 3 \cdot 2^{17} - 3 = 2 \cdot 2^{18} + 2^{17} - 3 = 2^{19} + 2^{17} - 3 $ which is not more than a million.

=> n = 19.

Let the first term of both series be $ a_1 $, and $ b_1 $, respectively, and the common difference be $ d_1 $, and $ d_2 $, respectively.

It is given that $ a_5 = b_9 $, which implies $ a_1 + 4d_1 = b_1 + 8d_2 $

=> $ a_1 - b_1 = 8d_2 - 4d_1 $ ..... Eq(1)

Similarly, it is known that $ a_{19} = b_{19} $, which implies $ a_1 + 18d_1 = b_1 + 18d_2 $

=> $ a_1 - b_1 = 18d_2 - 18d_1 $ ...... Eq(2)

Equating (1) and (2), we get:

=> $ 18d_2 - 18d_1 = 8d_2 - 4d_1 $

=> $ 10d_2 = 14d_1 $

=> $ 5d_2 = 7d_1 $

Since, $ d_1, d_2 $ are the prime numbers, which implies $ d_1 = 5, d_2 = 7 $.

It is also known that $ b_2 = 0 $, which implies $ b_1 + d_2 = 0 \Rightarrow b_1 = -d_2 = -7 $

Putting the value of $ b_1, d_1 $, and, $ d_2 $ in Eq(1), we get:

$ a_1 = 8d_2 - 4d_1 + b_1 = 56 - 20 - 7 = 29 $

Hence, $ a_{11} = a_1 + 10d_1 = 29 + 10 \cdot 5 = 29 + 50 = 79 $

The correct option is A

It is given that $ a_n = 13 + 6(n-1) $, which can be written as $ a_n = 13 + 6n - 6 = 7 + 6n $

Similarly, $ b_n = 15 + 7(n-1) $, which can be written as $ b_n = 15 + 7n - 7 = 8 + 7n $

The common differences are 6, and 7, respectively. The common difference of terms that exists in both series is l.c.m (6, 7) = 42

The first common term of the first two series is 43 (by inspection)

Hence, we need to find the mth term, which is less than 1000, and the largest three-digit integer, and exists in both series.

$ t_m $ = a + (m-1)d < 1000

=> $ 43 + (m-1)42 < 1000 $

=> $ (m-1)42 < 957 $

=> m-1 < 22.8 => m < 23.8 => m = 23

Hence, the 23rd term is $ 43 + 22 \times 42 = 967 $

It is given that $ \frac{1}{2} $, $ \frac{\log_3(2^x - 9)}{\log_3 4} $, and $ \frac{\log_5(2^x + \frac{17}{2})}{\log_5 4} $ are in an arithmetic progression.

$ \frac{\log_3(2^x - 9)}{\log_3 4} $ can be written as $ \log_4(2^x - 9) $, and $ \frac{\log_5(2^x + \frac{17}{2})}{\log_5 4} $ can be written as $ \log_4(2^x + \frac{17}{2}) $

Hence, $ 2 \log_4(2^x - 9) = \frac{1}{2} + \log_4(2^x + \frac{17}{2}) $

$ \frac{1}{2} $ can be written as $ \log_4 2 $.

Therefore,

=> $ 2 \log_4(2^x - 9) = \log_4 2 + \log_4(2^x + \frac{17}{2}) $

=> $ \log_4(2^x - 9)^2 = \log_4 2(2^x + \frac{17}{2}) $

=> $ (2^x - 9)^2 = 2(2^x + \frac{17}{2}) $

=> $ 2^{2x} - 18 \cdot 2^x + 81 = 2 \cdot 2^x + 17 $

=> $ 2^{2x} - 20 \cdot 2^x + 64 = 0 $

=> $ 2^{2x} - 16 \cdot 2^x - 4 \cdot 2^x + 64 = 0 $

=> $ 2^x(2^x - 16) - 4(2^x - 16) = 0 $

=> $ (2^x - 4)(2^x - 16) = 0 $

The values of $ 2^x $ can't be 4 (log will be undefined), which implies The value of $ 2^x $ is 16.

Therefore, the common difference is $ \log_4(2^x - 9) - \log_4 2 $

=> $ \log_4 7 - \log_4 2 = \log_4\left(\frac{7}{2}\right) $

The given sequence can be written as:

$ 1\left(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...\right) + \frac{1}{3}\left(\frac{1}{4} + \frac{1}{16} + ...\right) + \frac{1}{9}\left(\frac{1}{16} + \frac{1}{64} + ...\right) + ... $

We know that the sum of an infinite G.P. is $ \frac{a}{1-r} $, where a is the first term and r is the common ratio.

=> The first term = $ \frac{1}{1-\frac{1}{4}} = \frac{4}{3} $

=> The second term = $ \frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right) = \frac{1}{9} $

=> The third term = $ \frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right) = \frac{1}{108} $

Observing these three terms, we see that they are in G.P. with a common ratio of $ \frac{1}{12} $

=> Sum of this infinite G.P. = $ \frac{\left(\frac{4}{3}\right)}{1-\left(\frac{1}{12}\right)} = \frac{16}{11} $

The first series goes as follows:

46, 54, 62, 70, 78, 86, 94, 102,....

The second series goes as follows:

98, 102, 106, 110,...

The first common term is 102 (first term of the common terms) and the common difference between them will be lcm (4, 8) = 8

=> The required sequence is 102, 110, 118,..... (last term should be less than 468 (100th term of second series))

=> 102 + (n-1)(8) ≤ 498

=> n is less than or equal to 50.5 => n = 50

Using the summation of A.P. formula:

Required sum = $\frac{n}{2}(2a + (n - 1)d) = \frac{50}{2}(2 \times 102 + 49 \times 8) = 14900$

It is given,

$S_n = 2n^2 + n$

$S_{n-1} = 2(n - 1)^2 + (n - 1)$

$S_{n-1} = 2n^2 - 3n + 1$

$T_n = S_n - S_{n-1} = 2n^2 + n - 2n^2 + 3n - 1 = 4n - 1$

$T_n = 4n - 1$

The terms are 3, 7, 11, 15, 19, 23, 27,……

27 is the first term in the series divisible by 9.

27 is the 7th term.

Therefore, the least possible value of n is 7.

Given, the number of particles on day 1 = 100

On day 2, one out of every 2 articles produces another particle.

The number of particles on day 2 will be 100/2 , i.e. 50 particles

On day 3, one out of every 3 articles produces another particle.

The number of particles on day 3 will be (100+50)/3​, i.e. 50 particles

On day 4, one out of every 4 articles produces another particle.

The number of particles on day 4 will be (100+50+50)/4, i.e. 50 particles

Series will be 100, 50, 50, 50,....

It is given,

100 + (m-1)50 = 1000

m = 19

$a_1 ​+ a_2​ + ..... + a_N ​= 300_N$

$6_{a_1} + a_2 + ..... + a_N = 400_N$

$5_{a_1} ​= 100_N$

$a_1 ​= 20_N$

As the given sequence of numbers is non-decreasing sequence, N can take values from 2 to 15.

N is not equal to 1, if N = 1, then average of N numbers is 300 wouldn't satisfy.

Therefore, N can take values from 2 to 15, i.e. 14 values.

General term = $38 + (n-1)17 = 17n + 21 = 17(n+1) + 4 = 17k + 4$

Each term is in the form of $17k + 4$

Least 3-digit number in the form of $17k + 4$ is at $k = 6$, i.e. $106$

Highest 3-digit number in the form of $17k + 4$ is at $k = 58$, i.e. $990$

$106, 123, 140, \ldots, 990$

$990 = 106 + 17(n-1)$

$n = 53$

Sum = $\frac{53}{2}(106 + 990) = 53 \times 548$

Avg = $53 \times \frac{548}{53}$ = 548