# CAT Progressions Questions (with Notes) PDF

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Progressions and Series is one of the key topics in the CAT Quants section. CAT Progressions questions appear in the CAT and other MBA entrance exams almost every year. You can check out the CAT Progressions Questions from Previous years. In this article, we will look into some important CAT Progressions Questions (with Notes) PDF. These are a good source for practice; If you want to practice these questions, you can download this CAT Progressions Questions PDF, which is completely Free.

• CAT Progressions & Series Questions – Tip 1: Be thorough with all the basics of this topic. If you’re starting the prep, firstly understand the CAT Syllabus; CAT Progressions questions are not very tough, and hence must not be avoided.
• CAT Progressions & Series Questions – Tip 2: Understand the important formulas from Arithmetic Progressions (A.P), Geometric Progressions (G.P), and Harmonic Progression (H.P).
• Practice this CAT Progressions questions PDF. Learn all the major formulae from these concepts. You can check out the Important CAT Progressions & Series questions & Formulas PDF here.

Question 1:Â The number of common terms in the two sequences 17, 21, 25,â€¦, 417 and 16, 21, 26,â€¦, 466 is

a)Â 78

b)Â 19

c)Â 20

d)Â 77

e)Â 22

Solution:

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.

Question 2:Â The 288th term of the series a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,fâ€¦ is

a)Â u

b)Â v

c)Â w

d)Â x

Solution:

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
â€‹So, answer = x

Question 3:Â A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

a)Â 3

b)Â 4

c)Â 5

d)Â 6

e)Â 7

Solution:

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.

Question 4:Â If $a_1 = 1$ and $a_{n+1} = 2a_n +5$, n=1,2,….,then $a_{100}$ is equal to:

a)Â $(5*2^{99}-6)$

b)Â $(5*2^{99}+6)$

c)Â $(6*2^{99}+5)$

d)Â $(6*2^{99}-5)$

Solution:

$a_2 = 2*1 + 5$
$a_3 = 2*(2 + 5) + 5 = 2^2 + 5*2 + 5$
$a_4 = 2^3 + 5*2^2 + 5*2 + 5$

$a_{100} = 2^{99} + 5*(2^{98} + 2^{97} + … + 1)$
$= 2^{99} + 5*1*(2^{99} – 1)/(2-1) = 2^{99} + 5*2^{99} – 5 = 6*2^{99} – 5$

Question 5:Â What is the value of the following expression?
$(1/(2^2-1))+(1/(4^2-1))+(1/(6^2-1))+…+(1/(20^2-1)$

a)Â 9/19

b)Â 10/19

c)Â 10/21

d)Â 11/21

Solution:

$(1/(2^2-1))+(1/(4^2-1))+(1/(6^2-1))+…+(1/(20^2-1)$ = 1/[(2+1)*(2-1)] + 1/[(4+1)*(4-1)] + … + 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

Question 6:Â The value of $\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}$

a)Â $\frac{8}{1-x^8}$

b)Â $\frac{4x}{1+x^2}$

c)Â $\frac{4}{1-x^6}$

d)Â $\frac{4}{1+x^4}$

Solution:

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

Question 7:Â N the set of natural numbers is partitioned into subsets $S_{1}$ = $(1)$, $S_{2}$ = $(2,3)$, $S_{3}$ =$(4,5,6)$, $S_{4}$ = $(7,8,9,10)$ and so on. The sum of the elements of the subset $S_{50}$ is

a)Â 61250

b)Â 65525

c)Â 42455

d)Â 62525

Solution:

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 = $\frac{50}{2} (2\times 1226 + 49 \times 1 ) = 62525$

Question 8:Â If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

a)Â 2:3

b)Â 3:2

c)Â 3:4

d)Â 4:3

Solution:

The seventh term of an AP = a + 6d. Third term will be aÂ + 2d and second term will be aÂ + 16d. We are given that
$(a + 6d)^2 = (a + 2d)(a + 16d)$
=> $a^2$ + $36d^2$ +Â 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

Question 9:Â Let $a_1$, $a_2$,………….,Â  $a_{3n}$ be an arithmetic progression with $a_1$ = 3 and $a_{2}$ = 7. If $a_1$+ $a_{2}$ +…+ $a_{3n}$= 1830, then what is the smallest positive integer m such that m($a_1$+ $a_{2}$ +…+ $a_n$) > 1830?

a)Â 8

b)Â 9

c)Â 10

d)Â 11

Solution:

$a_{1}$ = 3 andÂ $a_{2}$ = 7. Hence, the common difference of the AP is 4.
We have been given that the sum up to 3n terms of this AP is 1830. Hence, $1830 = \frac{m}{2}[2*3 + (m – 1)*4$
=> 1830*2 = m(6 + 4m – 4)
=> 3660 = 2m + 4m$^2$
=> $2m^2 + m – 1830 = 0$
=> (m – 30)(2m + 61) = 0
=> m = 30 or m = -61/2
Since m is the number of terms so m 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($a_1$+Â $a_{2}$Â +…+ $a_n$) > 1830
=> 210m >Â 1830
=> m > 8.71
Hence, smallest integral value of ‘m’ is 9.

Question 10:Â Let $a_{1},a_{2},a_{3},a_{4},a_{5}$ be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with $2a_{3}$
If the sum of the numbers in the new sequence is 450, then $a_{5}$ is

Solution:

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 $a_{5} = 47 + 4 = 51$

Question 11:Â An infinite geometric progression $a_1,a_2,…$ has the property that $a_n= 3(a_{n+1}+ a_{n+2} + …)$ for every n $\geq$ 1. If the sum $a_1+a_2+a_3…+=32$, then $a_5$ is

a)Â 1/32

b)Â 2/32

c)Â 3/32

d)Â 4/32

Solution:

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 byÂ $\frac{a}{1-r}$ where a here isÂ $a_{n+1}$

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

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

Question 12:Â If $a_{1}=\frac{1}{2\times5},a_{2}=\frac{1}{5\times8},a_{3}=\frac{1}{8\times11},…,$ then $a_{1}+a_{2}+a_{3}+…+a_{100}$ is

a)Â $\frac{25}{151}$

b)Â $\frac{1}{2}$

c)Â $\frac{1}{4}$

d)Â $\frac{111}{55}$

Solution:

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

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

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

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

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

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

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

= $\frac{25}{151}$

Question 13:Â Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

a)Â $\frac{3}{6}$

b)Â $\frac{1}{6}$

c)Â $\frac{5}{2}$

d)Â $\frac{3}{2}$

Solution:

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

So,Â $r$ =Â $\frac{5}{2}$

Hence, option C is the correct answer.

Question 14:Â Let $\ a_{1},a_{2}…a_{52}\$ be positive integers such that $\ a_{1}$ < $a_{2}$ < … <Â $a_{52}\$. Suppose, their arithmetic mean is one less than arithmetic mean of $a_{2}$, $a_{3}$, ….$a_{52}$. If $a_{52}$= 100, then the largest possible value of $a_{1}$is

a)Â 48

b)Â 20

c)Â 23

d)Â 45

Solution:

Let ‘x’ be the average of all 52Â positive integersÂ $\ a_{1},a_{2}…a_{52}\$.

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

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

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

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

$a_{1}+51(x+1)$ =Â 52x

$a_{1}$ = x – 51.

We have to find out the largest possible value ofÂ $a_{1}$.Â $a_{1}$ will be maximum when ‘x’ is maximum.

(x+1) is the average of termsÂ $a_{2}$, $a_{3}$, ….$a_{52}$. We know thatÂ $a_{2}$ < $a_{3}$ < … <Â $a_{52}\$ andÂ $a_{52}$ = 100.

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

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

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

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

$\Rightarrow$ $x =Â 74$

Therefore,Â theÂ largest possible value ofÂ $a_{1}$ = x – 51 = 74 – 51 = 23.

Question 15:Â The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + …+ 95 x 99 is

Solution:

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,Â (4n+3) = 95 i.e. n = 23

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

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

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

$\Rightarrow$Â $80707$

Question 16:Â The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x â‰¥ z, then the minimum possible value of x is

Solution:

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

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

$\Rightarrow$ $x+y+z = 240$Â  … (1)

Also,Â Â $\dfrac{x+y+z+v+u}{5} = 75$

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

$\Rightarrow$Â $x+y+z+v+u = 375$

Substituting values from equation (1),

$\Rightarrow$Â $v+u = 135$

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

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

$\Rightarrow$Â $x+2y+zÂ = 270$

$\Rightarrow$Â $yÂ = 30$Â  Â (SinceÂ $x+y+z = 240$)

Therefore, we can say thatÂ $x+z = 240 – y = 210$. We are also given that x â‰¥ z,

Hence, $x_{min}$ = 210/2 = 105.

Question 17:Â Let $t_{1},t_{2}$,… be real numbers such that $t_{1}+t_{2}+â€¦+t_{n} = 2n^{2}+9n+13$, for every positive integer $n \geq 2$. If $t_{k}=103$, then k equals

Solution:

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

We can say that $t_{1}+t_{2}+â€¦+t_{k} = 2k^{2}+9k+13$Â  Â … (1)

Replacing k by (k-1) we can say that

$t_{1}+t_{2}+â€¦+t_{k-1} =Â 2(k-1)^{2}+9(k-1)+13$Â  Â … (2)

On subtracting equation (2) from equation (1)

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

$\Rightarrow$ $103 =Â 4k+7$

$\Rightarrow$ $k =Â 24$

Question 18:Â Let $a_1, a_2, …$ be integers such that
$a_1 – a_2 + a_3 – a_4 + …. + (-1)^{n – 1} a_n = n,$ for all $n \geq 1.$
Then $a_{51} + a_{52} + …. + a_{1023}$ equals

a)Â 0

b)Â 1

c)Â 10

d)Â -1

Solution:

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

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 andÂ $a_2, a_4, a_6….a_{2n}$ = -1

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

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

n=973

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

Question 19:Â If $(2n + 1) + (2n + 3) + (2n + 5) + … + (2n + 47) = 5280$, then whatis the value of $1 + 2 + 3 + .. + n?$

Solution:

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

Question 20:Â The number of common terms in the two sequences: 15, 19, 23, 27, . . . . , 415 and 14, 19, 24, 29, . . . , 464 is

a)Â 21

b)Â 20

c)Â 18

d)Â 19

Solution:

A:Â 15, 19, 23, 27, . . . . , 415

B:Â 14, 19, 24, 29, . . . , 464

Here the first common term = 19

Common difference = LCM of 5, 4=20

19+(n-1)20Â $\le\$ 415

(n-1)20Â $\le\$ 396

(n-1) $\le\$ 19.8

n=20

Question 21:Â If $a_1, a_2, ……$ are in A.P., then, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$ is equal to

a)Â $\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$

b)Â $\frac{n – 1}{\sqrt{a_1} + \sqrt{a_{n – 1}}}$

c)Â $\frac{n – 1}{\sqrt{a_1} + \sqrt{a_n}}$

d)Â $\frac{n}{\sqrt{a_1} – \sqrt{a_{n + 1}}}$

Solution:

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

Now,Â $\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$Â =Â $\frac{\sqrt{a_2} – \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} – \sqrt{a_1})}$Â  Â (Multiplying numerator and denominator byÂ $\sqrt{a_2} – \sqrt{a_1}$)

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

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

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

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

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

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

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

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

Question 22:Â If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

a)Â $(1003)^{15} + 6$

b)Â $(997)^{15} – 3$

c)Â $(997)2^{14} + 3$

d)Â $(1003)2^{15} – 3$

Solution:

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

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

Question 23:Â If $a_1 + a_2 + a_3 + …. + a_n = 3(2^{n + 1} – 2)$, for every $n \geq 1$, then $a_{11}$ equals

Solution:

11th term of series =Â $a_{11}$ = 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

Question 24:Â If $x_1=-1$ and $x_m=x_{m+1}+(m+1)$ for every positive integer m, then $X_{100}$ equals

a)Â -5050

b)Â -5151

c)Â -5051

d)Â -5150

Solution:

$x_1=-1$

$x_1=x_2+2$ =>Â $x_2=x_1-2$ = -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 isÂ $-\frac{n\left(n+1\right)}{2}$

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

Question 25:Â Let the m-th and n-th terms of a geometric progression be $\frac{3}{4}$ and 12. respectively, where $m < n$. If the common ratio of the progression is an integer r, then the smallest possible value of $r + n – m$ is

a)Â 6

b)Â 2

c)Â -4

d)Â -2

Solution:

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

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

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

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

Question 26:Â If $x_0 = 1, x_1 = 2$, and $x_{n + 2} = \frac{1 + x_{n + 1}}{x_n}, n = 0, 1, 2, 3, ……,$ then $x_{2021}$ is equal to

a)Â 4

b)Â 1

c)Â 3

d)Â 2

Solution:

$x_0=1$

$x_1=2$

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

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

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

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

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

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.

Question 27:Â The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), â€¦.. and so on. Then, the sum of the numbers in the 15th group is equal to

a)Â 6119

b)Â 6090

c)Â 4941

d)Â 7471

Solution:

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 beÂ $ax^2+bx+c$

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 beÂ $\left(15\right)^2-2\left(15\right)+2=197$

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=Â $\frac{29}{2}\left(197+225\right)\ =\ 6119$

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 isÂ $\left(14\right)^2=\ 196$, similarly for the 15th group this is :Â $\left(15\right)^2=\ 225$

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 :Â Â $\frac{n}{2}\cdot\left(first\ term\ +last\ term\right)$

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

6119.

Question 28:Â Three positive integers x, y and z are in arithmetic progression. If $y-x>2$ and $xyz=5(x+y+z)$, then z-x equals

a)Â 8

b)Â 12

c)Â 14

d)Â 10

Solution:

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)Â  = 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.

Question 29:Â For a sequence of real numbers $x_{1},x_{2},…x_{n}$, If $x_{1}-x_{2}+x_{3}-….+(-1)^{n+1}x_{n}=n^{2}+2n$ for all natural numbers n, then the sum $x_{49}+x_{50}$ equals

a)Â 200

b)Â 2

c)Â -200

d)Â -2

Solution:

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 $x_{49\ }+x_{50}\ =-2$

Question 30:Â Consider a sequence of real numbers, $x_{1},x_{2},x_{3},…$ such that $x_{n+1}=x_{n}+n-1$ for all $n\geq1$. If $x_{1}=-1$ then $x_{100}$ is equal to

a)Â 4849

b)Â 4949

c)Â 4950

d)Â 4850

Solution:

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

Considering

x1Â  Â = -1.Â  Â  Â  (1)

x2Â  Â = x1+1-1 = x1 + 0Â  Â  (2)

x3Â  Â = x2 + 2 – 1Â  =x2 + 1Â  Â  Â (3)

x4Â Â  = x3 + 3 – 1 = x3 + 2Â  Â  Â  Â  (4)

x100 = x99 + 98Â  Â  Â  (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 hasÂ a difference between the consecutive terms in an AP.

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

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,

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