CAT Progressions and Series Questions

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, 
$$2d=-4$$
$$d=-2$$
$$a=4$$

$$x_{100}=a+99d$$
$$x_{100}=4-198=-194$$

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. 

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

These are two infinite GPs when rearranged:
$$\left(\dfrac{1}{5}\right)^2+\left(\dfrac{1}{5}\right)^4+\left(\dfrac{1}{5}\right)^6+...-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2-...$$

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. 

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

$$\frac{\log_3(2^x-9)}{\log_34}$$ can be written as $$\log_4\left(2^x-9\right)$$, and $$\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$$ can be written as $$\log_4\left(2^x+\frac{17}{2}\right)$$ 

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

$$\frac{1}{2}$$ can be written as $$\log_42$$.

Therefore, 

=> $$2\log_4\left(2^x-9\right)=\log_42+\log_4\left(2^x+\frac{17}{2}\right)$$

=> $$\log_4\left(2^x-9\right)^{^2}=\log_42\left(2^x+\frac{17}{2}\right)$$

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

=> $$2^{2x}-18\cdot2^x+81=2\cdot2^x+17$$

=> $$2^{2x}-20\cdot2^x+64=0$$

=> $$2^{2x}-16\cdot2^x-4\cdot2^x+64=0$$

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

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

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 $$\log_4\left(2^x-9\right)-\log_42$$

=> $$\log_47-\log_42\ =\ \log_4\left(\frac{7}{2}\right)$$

The correct option is D

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

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

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

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

=> $$2x=y+z$$

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

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 $$b_1+d_2=0\ =>\ b_1=-d_2\ =\ -7$$

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

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

Hence, $$a_{11}=a_1+10d_1\ =\ 29+10\cdot5\ =\ 29+50\ =\ 79$$

The correct option is B

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 $$\dfrac{a}{1-r}$$, where a is the first term and r is the common ratio.

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

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

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

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

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

The first series goes as follows:

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

The second series goes as follows:

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) $$\le$$ 498

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

Using the summation of A.P. formula:

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

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

$$2^{19}+3\left(2^{18}-1\right)=2^{19}\ +\ 3\cdot2^{18}-3=2\cdot2^{19}+2^{18}-3=2^{20}+2^{18}-3$$, which is more than 1 million.

Let us check for n = 18

=> $$2^{18}+3\left(2^{17}-1\right)=2^{18}\ +\ 3\cdot2^{17}-3=2\cdot2^{18}+2^{17}-3=2^{19}+2^{17}-3$$ which is not more than a million.

=> n = 19.

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.

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

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

=> $$\left(m-1\right)42\ <\ 957$$

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

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

$$a_1+a_2+.....+a_N=300N$$

$$6a_1+a_2+.....+a_N=400N$$

$$5a_1=100N$$

$$a_1=20N$$

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.

It is given,

$$S_n=2n^2+n$$

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

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

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

$$T_n=4n-1$$

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.

The answer is option C.

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 $$\frac{100}{2}$$, 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 $$\frac{100+50}{3}$$, 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 $$\frac{100+50+50}{4}$$, i.e. 50 particles

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

It is given,

100 + (m-1)50 = 1000

m = 19

The answer is option A.

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,..........., 990

990 = 106 + 17(n-1)

n = 53

Sum = $$\frac{53}{2}\left(106+990\right)=53\times548$$

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

Sum of n terms in an A.P = $$\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)$$

$$A_n=\dfrac{n}{2}\left(6+\left(n-1\right)4\right)=n\left(2n+1\right)$$

$$\Sigma\ A_n=\Sigma\ n\left(2n+1\right)=2\Sigma\ n^2+\Sigma\ n=\ \dfrac{\ 2n\left(n+1\right)\left(2n+1\right)}{6}+\ \dfrac{\ n\left(n+1\right)}{2}$$

Substituting n = 25, we get

$$\dfrac{1}{25} \sum_{n=1}^{25} A_{n}$$ = $$\dfrac{1}{25}\left(\ \dfrac{\ 2\left(25\right)\left(25+1\right)\left(50+1\right)}{6}+\ \dfrac{\ 25\left(25+1\right)}{2}\right)$$

$$\dfrac{1}{25} \sum_{n=1}^{25} A_{n}$$ = $$\ 26\left(17\right)+13$$ = 455

The answer is option A. 

Given x, y, z are three terms in an arithmetic progression.

Considering x = a, y = a+d, z = a+2*d.

Using the given equation x*y*z = 5*(x+y+z)

 a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d)

=a*(a+d)*(a+2*d)  = 5*(3*a+3*d) = 15*(a+d).

= a*(a+2*d) = 15.

Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d are integers.

The common difference is positive and greater than 2.

Among the different possibilities are : (a=1, a+2d = 5), (a, =3, a+2d = 5), (a = 5, a+2d = 3), (a=15, a+2d = 1)

Hence the only possible case satisfying the condition is :

a = 1, a+2*d = 15.

x = 1, z = 15.

z-x = 14.

Given $$x_{n+1}\ =\ x_n\ +\ n\ -1$$ and x1 = -1.

Considering 

x1   = -1.      (1)

x2   = x1+1-1 = x1 + 0    (2)

x3   = x2 + 2 - 1  =x2 + 1     (3)

x4   = x3 + 3 - 1 = x3 + 2        (4)

x100 = x99 + 98      (100)

Adding the LHS and RHS for the hundred equations we have:

(x1+x2+......................x100) = (-1+0+.........98) + (x1+x2+...............x99)

Subtracting this we have :

(x1+...........x100) - (x1+............. x 99) = $$\frac{\left(98\cdot99\right)}{2}$$ - 1.

x100 = 4851 - 1 = 4850

Alternatively

$$x_1=-1$$

$$x_2=x_1+1-1=x_1=-1$$

$$x_3=x_2+2-1=x_2+1=-1+1=0$$

$$x_4=x_3+3-1=x_3+2=0+2=2$$

$$x_5=x_4+4-1=x_4+3=2+3=5$$

......

If we observe the series, it is a series that has a difference between the consecutive terms in an AP.

Such series are represented as $$t\left(n\right)=a+bn+cn^2$$

We need to find t(100).

t(1) = -1

a + b + c = -1

t(2) = -1

a + 2b + 4c = -1

t(3) = 0

a + 3b + 9c = 0

Solving we get,

b + 3c = 0

b + 5c = 1

c = 0.5

b = -1.5

a = 0

Now, 

$$t\left(100\right)=\left(-1.5\right)100+\left(0.5\right)100^2=-150+5000=4850$$

Now as per the given series :
we get $$x_1=1+2\ =3$$
Now $$x_1-x_2=\ 8$$
so$$x_2=-5$$
Now $$x_1-x_2+x_3\ =\ 15$$
so $$x_3\ =7$$
so we get $$x_n\ =\left(-1\right)^{n+1}\left(2n+1\right)$$
so $$x_{49}\ =\ 99$$ and $$x_{50}\ =-101$$
Therefore $$x_{49\ }+x_{50}\ =-2$$

$$x_0=1$$

$$x_1=2$$

$$x_2=\frac{\left(1+x_1\right)}{x_0}=\frac{\left(1+2\right)}{1}=3$$

$$x_3=\frac{\left(1+x_2\right)}{x_1}=\frac{\left(1+3\right)}{2}=2$$

$$x_4=\frac{\left(1+x_3\right)}{x_2}=\frac{\left(1+2\right)}{3}=1$$

$$x_5=\frac{\left(1+x_4\right)}{x_3}=\frac{\left(1+1\right)}{2}=1$$

$$x_6=\frac{\left(1+x_5\right)}{x_4}=\frac{\left(1+1\right)}{1}=2$$

Hence, the series begins to repeat itself after every 5 terms. Terms whose number is of the form 5n are 1, 5n+1 are 2... and so on, where n=0,1,2,3,....

2021 is of the form 5n+1. Hence, its value will be 2.

The first number in each group: 1, 2, 5, 10, 17.....

Their common difference is in Arithmetic Progression. Hence, the general term of the series can be expressed as a quadratic equation.

Let the quadratic equation of the general term be $$ax^2+bx+c$$

1st term = a+b+c=1

2nd term = 4a+2b+c=2

3rd term = 9a+3b+c=5

Solving the equations, we get a=1, b=-2, c=2.

Hence, the first term in the 15th group will be $$\left(15\right)^2-2\left(15\right)+2=197$$

We can see that the number of terms in each group is 1, 3, 5, 7.... and so on. These are of the form 2n-1. Hence, the number of terms in 15th group will be 29. Hence, the last term in the 15th group will be 197+29-1 = 225.

Sum of terms in group 15= $$\frac{29}{2}\left(197+225\right)\ =\ 6119$$

Alternatively,

The final term in each group is the square of the group number.

In the first group 1, second group 4, ............

The final element of the 14th group is $$\left(14\right)^2=\ 196$$, similarly for the 15th group this is : $$\left(15\right)^2=\ 225$$

Each group contains all the consecutive elements in this range.

Hence the 15th group the elements are:

(197, 198, ................................225).

This is an Arithmetic Progression with a common difference of 1 and the number of element 29.

Hence the sum is given by :  $$\frac{n}{2}\cdot\left(first\ term\ +last\ term\right)$$

$$\frac{29}{2}\cdot\left(197+225\right)$$

6119.

$$x_1=-1$$

$$x_1=x_2+2$$ => $$x_2=x_1-2$$ = -3

Similarly, 

$$x_3=x_1-5=-6$$

$$x_4=-10$$

.

.

The series is -1, -3, -6, -10, -15......

When the differences are in AP, then the nth term is $$-\frac{n\left(n+1\right)}{2}$$

$$x_{100}=-\frac{100\left(100+1\right)}{2}=-5050$$

Let the first term of the GP be "a" . Now from the question we can show that

$$ar^{m-1}=\frac{3}{4}$$    $$ar^{n-1}=12$$

Dividing both the equations we get $$r^{m-1-n+1}=\frac{1}{16}\ or\ r^{m-n}=16^{-1\ }or\ r^{n-m}=16$$

So for the minimum possible value we take Now give minimum possible value to "r" i.e -4 and n-m=2

Hence minimum possible value of r+n-m=-4+2=-2

We have, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$

Now, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$$ = $$\frac{\sqrt{a_2} - \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} - \sqrt{a_1})}$$   (Multiplying numerator and denominator by $$\sqrt{a_2} - \sqrt{a_1}$$)

= $$\frac{\sqrt{a_2} - \sqrt{a_1}}{({a_2} - {a_1}}$$

=$$\frac{\sqrt{a_2} - \sqrt{a_1}}{d}$$   (where d is the common difference)

Similarly, $$ \frac{1}{\sqrt{a_2} + \sqrt{a_3}}$$ = $$\frac{\sqrt{a_3} - \sqrt{a_2}}{d}$$ and so on.

Then the expression $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$

can be written as $$\ \frac{\ 1}{d}(\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_3}+..........................\sqrt{a_{n+1}} - \sqrt{a_{n}}$$

= $$\ \frac{\ n}{nd}(\sqrt{a_{n+1}}-\sqrt{a_1})$$ (Multiplying both numerator and denominator by n)

= $$\ \frac{n(\sqrt{a_{n+1}}-\sqrt{a_1})}{{a_{n+1}} - {a_1}}$$     $$(a_{n+1} - {a_1} =nd)$$

= $$\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$$

$$a_1 - a_2 + a_3 - a_4 + .... + (-1)^{n - 1} a_n = n$$

It is clear from the above equation that when n is odd, the co-efficient of a is positive otherwise negative.

$$a_1 - a_2 = 2$$

$$a_1 = a_2 + 2$$   

$$a_1 - a_2 + a_3 = 3$$

On substituting the value of $$a_1$$ in the above equation, we get

$$a_3$$ = 1

$$a_1 - a_2 + a_3 - a_4 = 4$$

On substituting the values of $$a_1, a_3$$ in the above equation, we get

$$a_4$$ = -1

$$a_1 - a_2 + a_3 - a_4 +a_5 = 5$$

On substituting the values of $$a_1, a_3, a_4$$ in the above equation, we get

$$a_5$$ = 1

So we can conclude that $$a_3, a_5, a_7....a_{n+1}$$ = 1 and $$a_2, a_4, a_6....a_{2n}$$ = -1

Now we have to find the value of $$a_{51} + a_{52} + .... + a_{1023}$$

Number of terms = 1023=51+(n-1)1

n=973

There will be 486 even and 487 odd terms, so the value of $$a_{51} + a_{52} + .... + a_{1023}$$ = 486*-1+487*1=1

The population of town at the beginning of 1st year = p

The population of town at the beginning of 2nd year = 3+2p

The population of town at the beginning of 3rd year = 2(3+2p)+3 = 2*2p+2*3+3 =4p+3(1+2)

The population of town at the beginning of 4th year = 2(2*2p+2*3+3)+3 = 8p+3(1+2+4)

Similarly population at the beginning of the nth year = $$2^{n-1}$$p+3($$2^{n-1}-1$$) = $$2^{n-1}\left(p+3\right)$$-3 

The population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be $$(2^{2034-2019})\left(1000+3\right)$$-3 = $$2^{15}\left(1003\right)$$-3

11th term of series = $$a_{11}$$ = Sum of 11 terms - Sum of 10 terms = $$3(2^{11 + 1} - 2)$$-3$$(2^{10 + 1} - 2)$$ 

= 3$$(2^{12} - 2-2^{11} +2)$$=3$$(2^{11})(2-1)$$= 3*$$2^{11}$$ = 6144

Let us first find the number of terms 

47=1+(n-1)2

n=24

24*2n+1+3+5+....47=5280

48n+576=5280

48n=4704

n=98

Sum of first 98 terms = 98*99/2

=4851

A: 15, 19, 23, 27, . . . . , 415

B: 14, 19, 24, 29, . . . , 464

Here the first common term = 19

Common difference = LCM of 5, 4=20

19+(n-1)20 $$\le\ $$ 415

(n-1)20 $$\le\ $$ 396

(n-1) $$\le\ $$ 19.8

n=20

It is given that $$t_{1}+t_{2}+…+t_{n} = 2n^{2}+9n+13$$, for every positive integer $$n \geq 2$$. 

We can say that $$t_{1}+t_{2}+…+t_{k} = 2k^{2}+9k+13$$   ... (1) 

Replacing k by (k-1) we can say that 

 $$t_{1}+t_{2}+…+t_{k-1} = 2(k-1)^{2}+9(k-1)+13$$   ... (2)

On subtracting equation (2) from equation (1)

$$\Rightarrow$$ $$t_{k} = 2k^{2}+9k+13 - 2(k-1)^{2}+9(k-1)+13$$

$$\Rightarrow$$ $$103 = 4k+7$$

$$\Rightarrow$$ $$k = 24$$

Let 'x' be the average of all 52 positive integers $$\ a_{1},a_{2}...a_{52}\ $$.

$$a_{1}+a_{2}+a_{3}+...+a_{52}$$ = 52x ... (1)

Therefore, average of $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$ = x+1

$$a_{2}+a_{3}+a_{4}+...+a_{52}$$ = 51(x+1) ... (2)

From equation (1) and (2), we can say that

$$a_{1}+51(x+1)$$ = 52x

$$a_{1}$$ = x - 51. 

We have to find out the largest possible value of $$a_{1}$$. $$a_{1}$$ will be maximum when 'x' is maximum. 

(x+1) is the average of terms $$a_{2}$$, $$a_{3}$$, ....$$a_{52}$$. We know that $$a_{2}$$ < $$a_{3}$$ < ... < $$a_{52}\ $$ and $$a_{52}$$ = 100.

Therefore, (x+1) will be maximum when each term is maximum possible. If $$a_{52}$$ = 100, then $$a_{52}$$ = 99, $$a_{50}$$ = 98 ends so on.

$$a_{2}$$ = 100 + (51-1)*(-1) = 50.

Hence,  $$a_{2}+a_{3}+a_{4}+...+a_{52}$$ = 50+51+...+99+100 = 51(x+1)

$$\Rightarrow$$ $$\dfrac{51*(50+100)}{2} = 51(x+1)$$

$$\Rightarrow$$ $$x = 74$$

Therefore, the largest possible value of $$a_{1}$$ = x - 51 = 74 - 51 = 23. 

S = 7 x 11 + 11 x 15 + 15 x 19 + ...+ 95 x 99

Nth term of the series can be written as $$T_{n} = (4n+3)*(4n+7)$$

Last term, (4n+3) = 95 i.e. n = 23

$$\sum_{n=1}^{n=23} (4n+3)*(4n+7)$$

$$\Rightarrow$$ $$\sum_{n=1}^{n=23}16n^2+40n+21$$

$$\Rightarrow$$ $$16*\dfrac{23*24*47}{6}+40*\dfrac{23*24}{2}+21*23$$

$$\Rightarrow$$ $$80707$$

Given that the arithmetic mean of x, y and z is 80.

$$\Rightarrow$$ $$\dfrac{x+y+z}{3} = 80$$

$$\Rightarrow$$ $$x+y+z = 240$$  ... (1)

Also,  $$\dfrac{x+y+z+v+u}{5} = 75$$

$$\Rightarrow$$ $$\dfrac{x+y+z+v+u}{5} = 75$$

$$\Rightarrow$$ $$x+y+z+v+u = 375$$

Substituting values from equation (1),

$$\Rightarrow$$ $$v+u = 135$$

It is given that u=(x+y)/2 and v=(y+z)/2.

$$\Rightarrow$$ $$(x+y)/2+(y+z)/2 = 135$$

$$\Rightarrow$$ $$x+2y+z = 270$$

$$\Rightarrow$$ $$y = 30$$   (Since $$x+y+z = 240$$) 

Therefore, we can say that $$x+z = 240 - y = 210$$. We are also given that x ≥ z, 

Hence, $$x_{min}$$ = 210/2 = 105. 

Let x = $$a$$, y = $$ar$$ and z = $$ar^2$$
It is given that, 5x, 16y and 12z are in AP.
so, 5x + 12z = 32y
On replacing the values of x, y and z, we get
$$5a + 12ar^2 = 32ar$$
or, $$12r^2 - 32r + 5$$ = 0
On solving, $$r$$ = $$\frac{5}{2}$$ or $$\frac{1}{6}$$

For $$r$$ = $$\frac{1}{6}$$, x < y < z is not satisfied.

So, $$r$$ = $$\frac{5}{2}$$

Hence, option C is the correct answer.

$$a_{100} = \frac{1}{ (3\times100 -1) \times (3\times100 + 2)}= \frac{1}{ 299 \times 302}$$

$$\frac{1}{2\times5} = \frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$$

$$\frac{1}{5\times8} = \frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$$

$$\frac{1}{8\times11} = \frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$$
....

$$\frac{1}{299\times302} = \frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$$

Hence $$a_{1}+a_{2}+a_{3}+...+a_{100}$$ = $$\frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$$ + $$\frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$$ + $$\frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$$ + ... + $$\frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$$

= $$\frac{1}{3} \times (\frac{1}{2} - \frac{1}{302})$$

= $$\frac{25}{151}$$

$$a_{1}$$ = 3 and $$a_{2}$$ = 7. Hence, the common difference of the AP is 4. If we assume, k=3n 
We have been given that the sum up to 3n terms of this AP is 1830. Hence, $$1830 = \frac{k}{2}[2*3 + (k - 1)*4$$
=> 1830*2 = k(6 + 4k - 4)
=> 3660 = 2k + 4k$$^2$$
=> $$2k^2 + k - 1830 = 0$$
=> (k - 30)(2k + 61) = 0
=> k = 30 or k = -61/2
Since k is the number of terms so k cannot be negative. Hence, must be 30
So, 3n = 30
n = 10
Sum of the first '10' terms of the given AP = 5*(6 + 9*4) = 42*5 = 210
m($$a_1$$+ $$a_{2}$$ +...+ $$a_n$$) > 1830
=> 210m > 1830
=> m > 8.71
Hence, smallest integral value of 'm' is 9.

Sum of the sequence of even numbers is $$2a_{3} + (2a_{3} - 2) + (2a_{3} - 4)$$ $$+ (2a_{3} - 6) + (2a_{3} - 8) = 450$$
=> $$10a_{3} - 20 = 450$$
=> $$a_{3} = 47$$
Hence $$a_{5} = 47 + 4 = 51$$

The seventh term of an AP = a + 6d. Third term will be a + 2d and second term will be a + 16d. We are given that
$$ (a + 6d)^2 = (a + 2d)(a + 16d)$$
=> $$ a^2 $$ + $$36d^2$$ + 12ad = $$ a^2 + 18ad + 32d^2 $$
=> $$4d^2 = 6ad$$
=> $$ d:a = 3:2$$
We have been asked about a:d. Hence, it would be 2:3

Let the common ratio of the G.P. be r.
Hence we have $$a_n= 3(a_{n+1}+ a_{n+2} + ...)$$

The sum up to infinity of GP is given by $$\frac{a}{1-r}$$ where a here is $$a_{n+1}$$

=> $$a_n= 3(\frac{a_{n+1}}{1-r})$$
=> $$a_n= 3(\frac{a_{n}\times r}{1-r})$$
=> $$ r = \frac{1}{4}$$
Now, $$a_1+a_2+a_2...+=32$$
=> $$\frac{a_1}{1-r} = 32$$
=> $$\frac{a_1}{3/4} = 32$$
=> $$a_1 = 24$$

$$a_5 = a_1 \times r^4$$
$$a_5 = 24 \times (1/4)^4 = \frac{3}{32}$$

The terms of the first sequence are of the form 4p + 13

The terms of the second sequence are of the form 5q + 11

If a term is common to both the sequences, it is of the form 4p+13 and 5q+11

or 4p = 5q -2. LHS = 4p is always even, so, q is also even.

or 2p = 5r - 1 where q = 2r.

Notice that LHS is again even, hence r should be odd. Let r = 2m+1 for some m.

Hence, p = 5m + 2.

So, the number = 4p+13 = 20m + 21.

Hence, all numbers of the form 20m + 21 will be the common terms. i.e 21,41,61,...,401 = 20.

Let x be in the front row.

So no. of children in next rows will be x-3,x-6,x-9,x-12,x-15,x-18,x-21....

Suppose there are 6 rows, then the sum is equal to x + x-3 + x-6 + x-9 + x-12 + x-15 = 6x - 45

This sum is equal to 630.

=> 6x - 45 = 630 => 6x = 585

Here, x is not an integer.

Hence, there cannot be 6 rows.

substituting 3,4...53 in the given function, we get
$$t_3 = \frac{3}{5}$$
$$t_4 = \frac{4}{6}$$
$$t_5 = \frac{5}{7}$$
$$t_6 = \frac{6}{8}$$

Multiplying the values, we get $$\frac{3}{5}*\frac{4}{6}*\frac{5}{7}*....\frac{52}{54}*\frac{53}{55} $$ which ultimately after cancellations give $$\frac{3*4}{54*55}=\frac{2}{495}$$

$$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$$

=> $$x = \sqrt{4+\sqrt{4-x}}$$

=> $$x^2 = 4 + \sqrt{4-x}$$

=>$$x^4 + 16 - 8x^2 = 4 - x$$

=> $$x^4 - 8x^2 + x +12 = 0$$

On substituting options, we can see that option C satisfies the equation.

Sum of the first 11 terms = 11/2 ( 2a+10d) 

Sum of the first 19 terms = 19/2 (2a+18d) 

=> 22a+110d = 38a+342d => 16a = -232d

=> 2a = -232/8 d = -29d

Sum of the first 30 terms = 15(2a+29d) = 0

According to given condition we have ,

n+6b =$$nr^6$$ and b=10.5n ,

63n+n = $$nr^6$$

$$r^6 = 64$$

r = 2

According to given conditions the terms are 81.33, -19, -100.33, -81.33, 19, 100.33, 81.33,-19,.. Hence the series repeats after every 6 terms . Also summation of these 6 terms is 0. Hence summation is 60002 terms will we sum of first 2 terms which is 62.33.

The sum of the 3rd and 15th terms is a+2d+a+14d = 2a+16d
The sum of the 6th, 11th and 13th terms is a+5d+a+10d+a+12d = 3a+27d
Since the two are equal, 2a+16d = 3a+27d => a+11d = 0
So, the 12th term is 0
 

1, 2, 3, 4,....n such that the sum is greater than 288
If n = 24, n(n+1)/2 = 12*25 = 300
So, n = 24, i.e. the 24th letter in the alphabet is the letter at position 288 in the series
​So, answer = x

For the given problem ,

$$\sum {n(n+1)/2} = 8436 $$ which is 

$$\sum {n^2/2} + \sum{n/2} = 8436 $$ which is equal to

n*(n+1)(2n+1)/12 + n*(n+1)/4 = 8436 , solving we get n=36.

Solving the equation might be lengthy. you can substitute the values in the options to arrive at the answer. 

Let the numbers be $$a_1,a_2....a_n.$$

Since the numbers are positive,

$$AM\geq GM$$

$$\frac{a_1+a_2+a_3....+a_n}{n}\geq (a_1*a_2....*a_n)^{1/n}$$

$$a_1+a_2+a_3....+a_n \geq n$$

Let there only be 2 questions.

Thus there are $$2^{2-1}$$ = 2 students who have done 1 or more questions wrongly, 2$$^{2-2}$$ = 1 students who have done all 2 questions wrongly .

Thus total number of wrong answers = 2 + 1 = 3= $$2^n - 1$$.

Now let there be 3 questions. Then j = 1,2,3

Number of students answering 1 or more questions incorrectly = 4

Number of students answering 2 or more questions incorrectly = 2

Number of students answering 3 or more questions incorrectly = 1

Total number of incorrect answers = 1(3)+(2-1)*2+(4-2)*1 = 7 = $$2^3-1$$

According to the question , the total number of wrong answers = 4095 = $$2^{12} - 1$$.

Hence Option A.

Let x = 1, so, $$X_0$$ = 1
$$X_1$$ = -1
$$X_2$$ = -1
$$X_3$$ = 1
$$X_4$$ = 1
$$X_5$$ = -1
$$X_6$$ = -1
So, $$X_n$$ need not be positive when n is even, $$X_n$$ need not be positive when n is odd, $$X_n$$ need not be negative when n is even. So, none of the first three options are correct.

Let $$S = 2+5x+9x^2+....$$
$$S*x = 2x+5x^2+9x^3+...$$
$$S(1-x) = 2+3x+4x^2+...$$
$$S(1-x)*x = 2x+3x^2+4x^3+...$$
$$S(1-x)(1-x) = 2+x+x^2+x^3+... = 2+x/(1-x)$$
So, $$S = [2(1-x) + x]/(1-x)^3 => S = (2-x)/(1-x)^3$$

If the child adds all the numbers from 1 to 34, the sum of the numbers would be 1+2+3+...+34 = 34*35/2 = 595
Since the child got the sum as 575, he would have missed the number 20.

January 1, 1950 to December 31, 1959 is a period of 10 years or 20 half years.
The person X after 1st year gets Rs. 300 in next year he gets Rs. 330 and so on.
So his earning is in AP with 10 300+330+360+...

Similarly earning of Y is in AP with 20 terms 200+215+230+245.... .

So, the total earnings of X equals 12*(300+330+....10 terms) = 52200
The total earnings of Y equals 6*(200+215+230+...20 terms) = 41100

So, the total earnings of the two equals 52200+41100 = 93300

It is given that in a Fibonacci sequence, from the third term on wards, each term in the sequence is the sum of the previous two terms in that sequence.

Let x and y be the 1st and 2nd term respectively.

3rd term = x+y

4th term = x+2y

5th term = 2x+3y

6th term = 3x+5y

7th term = 5x+8y

We know that difference of the squares of 6th and 7th terms is 517 = 47*11 .

And $$a^2-b^2=(a+b)(a-b)$$.

Applying above formula we get (8x+13y)(2x+3y) = 47*11.

So only possible way is (8x+13y)=47 and

2x+3y=11 .

Solving we get x=1 and y=3 .

Using the concept that every term is the sum of the previous two terms, as used in the beginning of the solution, we get 10th term as 21x+34y, which gives 10th term as 123.

$$a_2 = 2*1 + 5$$
$$a_3 = 2*(2 + 5) + 5 = 2^2 + 5*2 + 5$$
$$a_4 = 2^3 + 5*2^2 + 5*2 + 5$$ 
...
$$a_{100} = 2^{99} + 5*(2^{98} + 2^{97} + ... + 1)$$
$$= 2^{99} + 5*1*\dfrac{(2^{99} - 1)}{(2-1)} = 2^{99} + 5*2^{99} - 5 = 6*2^{99} - 5$$

$$\dfrac{1}{(2^2-1)}+\dfrac{1}{(4^2-1)}+\dfrac{1}{(6^2-1)}+...+\dfrac{1}{(20^2-1)}$$ 

= $$\dfrac{1}{(2+1)(2-1)}+\dfrac{1}{(4+1)(4-1)}+..+\dfrac{1}{(20+1)(20-1)}$$

= 1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/(19*21) 

=1/2 * ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... +1/19 - 1/21)

=1/2 * (1 - 1/21) = 10/21

We will give an example to disprove each of the three options A, B and C and hence, the correct answer will be option D.

Initially, if the least integer in S1 is -20 and the greatest integer in S2 is 50, then after the doing the operations mentioned in the question, the greatest integer in S1 is not greater than 50. Hence option A is false.

G is in S2 as per the given information => Option B is false.

If S1 contains numbers from 1 to 24 and S2 contains numbers from 25 to 50. Then G = 50 and L = 1. If all the numbers of S1 change in sign, G will remain 50 and will be in S2 while L will be -24 and will be in S1.

Hence, none of the statements is true always.

We know that $$a_{24}$$ is less than $$a_{25}$$.

So, even if $$a_{25}$$ replaces $$a_{24}$$, the ascending order still exists in S1.

But, $$a_{25}$$ is less than $$a_{26}$$. Hence, the descending order does not exist in S2 anymore.

For the least element L in $$S1$$ to be greater than the greatest element or equal to G in $$S2$$, the number that is added to L cannot be less than G - L.

Assume the first series as a,b,a/2,c,a+20
and second series as x1,x2,x3,x4
x1=b-a, x2= a/2-b, x3=c-a/2, and x4=a+20-c
x2-x1=30 => 3a-4b=60
and x4-x3=30 => 3a-4c=20
and x4-x2=60 => a-2c+2b=80
Solving we get, a=100, b=60, and c=70
S1= 100,60,50,70,120

S2 = -40, -10, 20, 50

Assume the first series as a,b,a/2,c,a+20
and second series as x1,x2,x3,x4
x1=b-a, x2= a/2-b, x3=c-a/2, and x4=a+20-c
x2-x1=30 => 3a-4b=60
and x4-x3=30 => 3a-4c=20
and x4-x2=60 => a-2c+2b=80
Solving we get, a=100, b=60, and c=70
S1= 100,60,50,70,120
S2 = -40, -10, 20, 50

Assume the first series as a,b,a/2,c,a+20
and second series as x1,x2,x3,x4
x1=b-a, x2= a/2-b, x3=c-a/2, and x4=a+20-c
x2-x1=30 => 3a-4b=60
and x4-x3=30 => 3a-4c=20
and x4-x2=60 => a-2c+2b=80
Solving we get, a=100, b=60, and c=70
S1= 100,60,50,70,120

S2 = -40, -10, 20, 50

Difference = 20

Assume the first series as a,b,a/2,c,a+20
and second series as x1,x2,x3,x4
x1=b-a, x2= a/2-b, x3=c-a/2, and x4=a+20-c
x2-x1=30 => 3a-4b=60
and x4-x3=30 => 3a-4c=20
and x4-x2=60 => a-2c+2b=80
Solving we get, a=100, b=60, and c=70
S1= 100,60,50,70,120

S2 = -40, -10, 20, 50

Avg. = $$(100+60+50+70 +120) \div 5 = 80$$

Assume the first series as a,b,a/2,c,a+20
and second series as x1,x2,x3,x4
x1=b-a, x2= a/2-b, x3=c-a/2, and x4=a+20-c
x2-x1=30 => 3a-4b=60
and x4-x3=30 => 3a-4c=20
and x4-x2=60 => a-2c+2b=80
Solving we get, a=100, b=60, and c=70
S1= 100,60,50,70,120

S2 = -40, -10, 20, 50
Sum = 20

Suppose consecutive odd integers are: (a-2), a, (a+2)

Hence, 3a-6 = 2(a+2) + 3 => a=13

a+2 = 15

Given series can be written as:

$$\sum_{n=1}^{100} (\frac{1}{n\times (n+1)})$$

or $$\sum_{n=1}^{100} (\frac{(n+1)-n}{n\times (n+1)})$$

or $$\sum_{n=1}^{100} (\frac{1}{n} - \frac{1}{n+1})$$

After putting values of n from 1 to 100, all terms will cancel out, only first and last terms will be there

i.e. $$1-\frac{1}{101}$$

or $$\frac{100}{101}$$

$$\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$$
or $$\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$$
or $$\frac{4}{1-x^4}+\frac{4}{1+x^4}$$
or $$\frac{8}{1-x^8}$$

$$\frac{-p}{1-p}$$ can be compared with sum of an infinite G.P. series i.e. $$\frac{a}{1-r}$$ (where a is first term and r is common ratio)
Hence here a=(-p)
and r = p
So k^th term will be = $$(-p) (p)^{(k-1)}$$

According to given question $$S_{50}$$ will have 50 terms 
And its first term will be 50th number in the series 1,2,4,7,.........$$T_{50}$$
$$T_1 = 1$$
$$T_2 = 1+1$$
$$T_3 = 1+1+2$$
$$T_4 = 1+1+2+3$$
$$T_n = 1+(1+2+3+4+5....(n-1))$$
       = $$1+\frac{n(n-1)}{2}$$
So $$T_{50} = 1+1225 = 1226$$
Hence $$S_{50} = (1226,1227,1228,1229........)$$
And summation will be = $$\frac{50}{2} (2\times 1226 + 49 \times 1 ) = 62525$$

Side of first square = 8cm. 

Side of second square made by joining mid-points of first square = $$\frac{8}{\sqrt2}$$

Similarly side of third square =  $$\frac{8}{\sqrt2\times\sqrt2}$$ and so on.

Now summation of areas will be $$8^2+(\frac{8}{\sqrt2})^2+(\frac{8}{\sqrt2\times\sqrt2})^2$$ .........

or $$8^2 ( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}.....)$$

or $$64 \times (\frac{1}{1-\frac{1}{2}})$$ (As we know sum of an infinite G.P. is $$\frac{a}{1-r}$$ where a is first term and r is common ratio)

or 128 sq. cm.