CSAT Problems on Sequences - Ananta IAS Academy

CSAT 12-09-2020

Problems on Sequences

1)Building a positive attitude

There was a man who made a living selling balloons at a fair. He had all colors of balloons, including red, yellow, blue, and green. Whenever business was slow, he would release a helium-filled balloon into the air and when the children saw it go up, they all wanted to buy one. They would come up to him, buy a balloon, and his sales would go up again. He continued this process all day. One day, he felt someone tugging at his jacket. He turned around and saw a little boy who asked, “If you release a black balloon, would that also fly?” Moved by the boy’s concern, the man replied with empathy, “Son, it is not the color of the balloon, it is what is inside that makes it go up.”

The same thing applies to our lives. It is what is inside that counts. The thing inside of us that makes us go up is our attitude.

William James of Harvard University said, “The greatest discovery of my generation is that human beings can alter their lives by altering their attitudes of mind.”

Sequence

A set of numbers arranged in a definite order according to some definite rule is called sequence.

i.e A sequence is a set of numbers written in a particular order.

Now take a sequence

m1, m2, m3, m4, . . . . . . . . , mn

Here first term in a sequence is m1, the second term m2, and so on. With this same notation, nth term in the sequence is mn.

Examples of sequence

1. Take the numbers 1, 5, 9, 13, 17, . . . . . .

In the above numbers seem to have a rule. It is started with the a number 1, and add a number 4 to previous number to obtained successive number

2. Take the numbers 1, 5, 25, 125, 625 . . . . . .

In the above numbers seem to have a rule. It is started with the a number 1, and multiplying with a number 5 to previous number to obtained successive number

3. Take the numbers 1, 8, 27, 64, 125, 216 . . . .

This is the sequence of cube of numbers.

4. Take the numbers 2 , −2, 2, −2, 2, −2, . . .

In the above sequence of numbers alternating between 2 and −2.

In above all examples, the dots written at the end indicate that we must consider the sequence as an infinite sequence.

5. The numbers 1, 9, 17, 25, 33

In the above example having form a finite sequence containing just five numbers.

6. Take numbers 1, 3, 5, 7, 9, 11, . . . . . . . n

In the above sequence, the last number is mentioned as ‘n’. So it is finite sequence

Series

The sum of the terms of a sequence is called a series

i.e A series is obtained from a sequence by adding all the terms together.

Take a sequence m1, m2, m3, m4 . . . . . . . . , mn

The series we obtain from this is m1 + m2+ m3 + m4 + . . . . . . . . + mn

Finite series and Infinite series

The series 1 + 9 + 17 + 25 + 33 is called finite series

The series 1 + 9 + 17 + 25 + 33+ . . . . . . . . . . . . is called infinite series

Sequence and series examples

1. A sequence is given by the formula mn = 2n + 3, for n = 1, 2, 3, . . .. now find first four terms of this sequence.

Solution: nth term of the sequence is mn = 2n + 3

First term of the sequence is = 2 ( 1)+3 = 5 ( substitute n = 1)

Second term of the sequence is = 2 ( 2 )+3 = 7 ( substitute n = 2)

Third term of the sequence is = 2 ( 3 )+3 = 9 ( substitute n = 3)

Fourth term of the sequence is = 2 ( 4)+3 = 11 ( substitute n = 4)

2. A sequence is given by mn = 2/(n+2), for n = 0, 1, 2, 3, . . .then find 8th term?

Solution: nth term of the sequence is mn = 2/(n+2)

Now 8th term of the sequence is (1.e n= 7) = 2/9

3. Find first three terms of the following sequence, beginning with n = 0,1 & 2. mn = 2/(n+2)!

Solution: nth term of the sequence is mn = 2/(n+2)!

Now 1st term of the sequence is (1.e n= 0) = 2/2! = 2/2 = 1

2nd term of the sequence is (1.e n= 1) = 2/3! = 2/6 = 1/3

3rd term of the sequence is (1.e n= 2) = 2/4! = 2/24 = 1/12

So the sequence is 1 , 1/3 , 1/12

Sequence following specific patterns are called progressions.

Progressions are of three types. They are

1. Arithmetic Progression

2. Geometric Progression

3. Harmonic Progression

Arithmetic Progression

Arithmetic Progression can be defined as, a sequence of numbers is obtained by adding a fixed number “d” to the preceding term except the first term.

i.e A list of numbers X1, X2, X3, X4 . . . . . . . is an A.P, If the differences X2-X1, X3-X2, X4-X3, . . . . given the same value

Each of the numbers in the list of an arithmetic progression is called a term of that A.P.

The general form of an A.P is a, a+d, a+2d, a+3d . . . . . . ( Here “d” is the common difference and “a” is the first term )

Common Difference

The difference between two succeeding terms of an Arithmetic Progression is called the common difference. The difference value can be positive,negative or zero.

Examples:

1) Rainfall of the last week ( in mm) 12, 10, 08, 06, 04, 02, 0

2) Temperature record for the last week ( in Deg.C) 30, 31, 32, 33, 34, 35, 36

3) 2, 4, 6, 8, 10, . . . . . . . . . . . .

4) 0, -5, -10, -15, -20, . . . . . . . . . . .

Finite and infinite arithmetic Progressions

In the above examples no. 1 & 2 having finite number of terms. Such arithmetic Progressions is called finite A.P. In the same way above examples no. 3 & 4 having infinite number of terms. So they are called infinite arithmetic Progressions

Formula for finding the nth term or general term of AP formula

Let t1, t2, t3, t4, . . . . . be an AP, Here first term ‘t1‘ is “a and common difference is “d”

Then,

First term a1 = a = a + (1 -1 ) d

Second term a2 = a + d = a+ ( 2-1)d

Third term a3 = a2 + d = a + ( 2-1)d +d = a + (3 -1) d

Fourth term a4 = a3 + d = a + (3 -1) d +d = a + (4 -1) d

Firth term a5 = a4 + d = a + (4 -1) d +d = a + (5 -1) d

Similarly use same pattern for nth term

an = a+ (n -1) d

an is also called general term of the A.P

Using the above formula we can find different terms of the A.P

General notation of nth terms of an Arithmetic progression

Tn = a + ( n-1) d

Here Tn = nth of an A.P ; a = first term of the A.P ; d = common difference

If number of terms in an A.P is “ r” and last term is “L” then,

L = a + ( r – 1)d

Formula for finding the sum to first nth term an A.P

Let a, a+d, a+2d, a+3d, . . . . . . . . . be an AP, Here first term “a” and common difference is “d‘

The nth term of this A.P is an = a + ( n-1) d

Let Sum of “n” terms of an A.P is ” Sn“

Sn = a + (a+d) + (a+2d) + ( a+3d) + . . . . . . . . . + [ a + ( n-1) d ]

Now write Sn from last term of A.P

Sn = [ a + ( n-1) d ] + [ a + ( n-2) d ] + [ a + ( n-3) d ] + . . . . . . . . . . . + a

Adding Sn +Sn

⇒ 2Sn = [ 2a + ( n-1) d ] + [ 2a + ( n-1) d ] + [ 2a + ( n-1) d ] +. . . . . . . . . . .+ [ 2a + ( n-1) d ] ( n times)

⇒ 2Sn = n [ 2a + ( n-1) d ]

⇒ $Sn \ = \frac{n}{2} \ [2a \ + \ (n-1)d)]$

⇒ Sn = (n/2) [ a + a + ( n-1) d ] = (n/2) ( first term + nth term)

⇒ Sn = (n/2) ( a + an)

Problems on Percentages and Progressions:

1. In market research conducted by an automobile company, the depreciation in the resale value is found to be 15% for one year old model, 13.5% for two years old model, 12% for three years old model and so on. If a luxury car made by the company was bought at Rs.8,00,000 ten years ago, then what will be the resale price of that car?

a) Rs 2,80,000

b) Rs 1,40,000

c) Rs 3,20,000

d) None of these

Answer: b)

Explanation:

It is clear from the question that the depreciation in resale is in form of an A.P.

15, 13.5, 12….

for which a = 15 and d = – 1.5

After ten years, the percentage depreciation will be:

Vedic Mathematics Principle:

ऊर्ध्वतिर्यग्भ्याम्

Urdhva Tiryagbhyam Vertically and Crosswise

Multiplication of 2 digit number tricks :

In this process having 3 steps. for easy understanding purpose, here explained with alphabets as a number.

Take ab x cd

Let us go with one example 47 X 56

Step 1 : ( b x d ) = 7 x 6 = 42. Take ” 2 ” for answer and “4” to be carry over to next step ( Ans : 2).

Step 2 : [ (a x d) + (b x c) ] + add number if any carry over from previous step

i.e [ (4 x 6) + (5 x 7) ] + 4 = 63. Take ” 3 ” for answer and “6” to be carry over to next step ( Ans : 32)

Step 3 : ( a x c ) + add number if any carry over from previous step

i.e 4 x 5 =20 + 6 = 26 Take ” 26 ” ( Ans : 2632)

So final answer is 2632

2. Alex bought an item from a sale offering 15% discount on all items. How much percentage more than the cost price should he sell the same item to earn 19 % profit from the list price?

a) 66.66%

b) 32%

c) 40%

d) 28.57%

Answer: c)

Explanation: Let the original list price be Rs 100. According to question, cost price will be Rs 85.

Also, Re-Selling price = 100 + 19 = 119

Problems on Ratio and Proportion

3. Two friends work in same company and ratio of their salaries is 3 : 5. First friend gets promoted after a year and his salary is increased by one third of the salary of his friend. Find the ratio of their current salaries.

a) 11 : 15

b) 2 : 3

c) 14 : 15

d) None of these

Answer: c)

Explanation:

4. Amit got thrice as many marks in Maths as in English. The proportion of his marks in Maths and History is 4:3. If his total marks in Maths, English and History are 250, what are his marks in English?

a) 120

b) 90

c) 40

d) 80

Answer: c)

Explanation:

Time and Work

5. A can finish solving an exercise in 2 hours, B in 3 hours and C in 4 hours respectively. In how many hours will all three complete 100 exercises, considering that no two people work on same exercise?

Answer: a)

Explanation: LCM of 2, 3 and 4 = 12

In 12 hours, A will finish 6 exercises

B will finish 4 exercises

C will finish 3 exercises

i.e., in 12 hours they will finish 13 exercises

So, in 84 hours they will finish 91 exercises

In next 9 hours, A will finish 4 exercises

B will finish 3 exercises

C will finish 2 exercises

So, they will complete 100 exercises in 93 hours.

Note: Since, they cannot share each-others work so B will take completely 9 hours to make 3 shawls, even when A and C stay idle for4 the last 1 hour till B completes his last exercise.

6. A and B, working together, complete a task in 10 days. Had A worked at half of his efficiency and B at five times of his efficiency, it would have taken them 50% of the scheduled time to finish the job. In how many days can B complete the job working alone?

a) 12

b) 24

c) 15

d) 30

Answer: d)

Explanation:

LCM and HCF

7. A sandwich is made from 2 bread slices. Three loafs of bread having weight 637gm, 581gm and 553gm respectively, are sliced evenly such that each slice has maximum possible weight. How many sandwiches can be made using the available bread slices?

a) 253

b) 252

c) 126

d) None of these

Answer: c)

Explanation:

Clock

8. A clock is set right at 10 a.m. The clock loses 15 minutes in 24 hours. What will be true time when clock indicates 10 p.m on the third day?

a) 11:20 PM

b) 11: 35 PM

c) 11:00 PM

d) None of these

Answer: c)

Explanation:

Mixture and Alligation

9. A jar contains a mixture of 2 liquids P and Q in the ratio 4:1. When 10 litres of the mixture is taken out and 10 litres of liquid Q is poured into the jar, the ratio becomes 2:3. How many litres of the liquid P was contained in the jar?

a) 17 litres

b) 16 litres

c) 18 litres

d) 15 litres

Answer (b)

Explanation:

10. Barrel A contains a mixture of kerosene and oil in the ratio 5:2, and Barrel B contains the mixture of oil and kerosene in the ratio 6: 7. In what ratio should these mixtures be mixed to obtain a new mixture containing kerosene and oil in the ratio 8: 5?

a) 4 : 3

b) 3 : 4

c) 5 : 6

d) 7 : 9

Answer: d)

Explanation:

Problems on Permutations:

11. How many words can be made from the word BING BING BRING in which vowels occupies even positions?

a) 336000

b) 85360

c) 113600

d) None of the above

Answer: a)

Explanation:

Q)If A + B = 2C and C + D = 2A, then

A) A + C =B+ D	B) A + C =2D
C) A + D = B + C	D) A + C = 2BExplanation:A+B-2A=2C-C-DB-A=C-DB+D=A+C

Level of Difficulty-I

1. A one rupee coin is placed on a table. The maximum number of one rupee coins which can be placed around it with each one of them touching it and two others is

(a) 8

(b) 6

(d) 4

Level of Difficulty-II

2. At a certain fast food restaurant, Brian can buy 3 burgers, 7 shakes, and one order of fries for Rs. 120 exactly. At the same place it would cost Rs. 164.5 for 4 burgers, 10 shakes, and one order of fries. How much would it cost for an ordinary meal of one burger, one shake, and one order of fries?

(a) Rs. 31

(b) Rs. 41

(d) Cannot be determined

3. Three pencils, four erasers and two sharpeners together cost Rs. 10.50. Two pencils, one eraser and three sharpners together cost Rs. 9.50. How much does each one eraser, one pencil, one sharpener cost?

(a) Rs. 3

(b) Rs. 4

(d) Can’t be determined

4. How many three-digit numbers can be generated from 1,2,3,4,5,6,7,8,9 such that the digits are in ascending order ?

(a) 80

(b) 81

(d) 84

Level of Difficulty-III

5. Four friends start from four towns, which are at the four corners of an imaginary rectangle. They meet at a point which falls inside the rectangle, after travelling the distance of 40 m, 50 m and 60m. The maximum distance that the fourth could have travelled is approximately:

(a) 67 m

(b) 52 m

(d) Can’t be determined

1. In a birthday party, each child eats at most 5 toffees. No child eats less than three biscuits. Considering all children, there are more toffees than biscuits, more biscuits than chocolates and more chocolates than number of children. Find the minimum number of children present at the party?

(a) 2

(b) 3

(d) 7

Answer (b)

Explanation:

Let p, q, r and s be the number of toffees, biscuits, chocolates and children respectively.

According to question, p > q > r > s

Minimum number of children can be 2 i.e. s > 2, accordingly, p≤4

But according to question, p≥5, therefore s cannot be taken as 2.

Considering s = 3, accordingly p≤6, which can satisfy p≥5

Thus, minimum number children at party will be 3.

Category: Puzzle

2. In a cinema hall, 8 friends, H, I, J, K, L, M, N, O are sitting in a straight line according to the following rules:

J is the neighbour of K.
H is just next to the left of L.
I is in one of the two seats in the middle.
M is the only one sitting between I and L.

If N is not sitting at the extreme end, then who occupies the extreme right seat?

(a) H

(b) J

(d) None of these

Answer (d)

Explanation:

The possible solutions are: H N/O L M I O/N J/K K/J

Thus, Either J or K occupies the extreme right seat.

Category: Puzzle

3. Which does not belong in this sequence?

(a) C

(b) B

(d) E

Answer (b)

Explanation:

In the others the long hand moves 45° clockwise and the short hand moves 90° clockwise.

Category: Figure matrix

4. Consider the following matrix:

What is the missing number in the given matrix?

(a) 15

(b) 18

(d) 19

Answer (C)

Explanation:

The value at each corner of the diagram equals the sums of difference of the numerical values of the letters in the boxes adjacent to the corner in clockwise direction starting from NRG.

Category: Series

5. Which number replaces the question mark ?

(a) 11.5

(b) 12.25

(d) 9.5

Answer (d)

Explanation:

Category: Puzzle

6.Find the number of triangles in the given figure?

(a) 32

(b) 36

(d) None of these

Answer (c)

7. How many colours are required at the minimum to paint the given figure such that no two adjacent regions have the same colour? (Regions sharing a vertex should not be considered as adjacent regions)?

(a) 2

(b) 3

(d) 5

Answer (b)

Explanation:

Category: Puzzle

8. A dice is given in (X), when it is folded which of I, II, II and IV will be a possible figure?

(a) I, II and III

(b) I and III

(d) I, II and IV

Answer (b)

9.In the following question, select a figure from amongst four alternatives, which figure placed in blank space of figure X would complete the pattern.

Answer (a)

10. What will come in place of the question mark?

(a) 4

(b) 7

(d) 3

Answer (a)

Explanation:

Each number in the segment at the bottom is the sum of the four numbers in the sections either side.

Thus 5 + 7 +? + 3 = 19.

? = 4