Number Series

Two Stage Type Series

The Two Stage Type Series is a series in which the difference between consecutive terms forms a series of its own. In such series, we have a parent series whose terms have a regular difference between them. The difference between each consecutive term of the parent series forms an arithmetic progression of its own. Here we will see what an arithmetic progression is and how we get a Two-Stage Type Series from it. Let us proceed.

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Two Stage Type Series

Let us first see how such a series looks. For example, consider the following series of numbers: 1, 3, 6, 10, 15. As you can see the rule is that a certain number is added to the previous term to get the subsequent term of the series. For example, we can write the series as 1 (1), 3 (1 + 2), 6 (3 + 3), 10 (6 + 4), 15 (10 + 5). Therefore, we can say that each term can be got from the previous term by adding a regular term to it. If you take the numbers that when we add to form the sequence we get the following sequence 2, 3, 4, 5.

All of the terms in this sequence have a constant difference. Such a sequence is known as the Arithmetic Progression or the A.P. So we define the Two-Stage Type Series as a series in which the difference between terms forms an arithmetic progression. Let us see Arithmetic Progression in a bit of a detail.

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A.P. Or The Arithmetic Progression

A sequence of numbers in which the difference between consecutive numbers is a constant number forms an Arithmetic Series or an A.P. In an A.P. the difference between any two consecutive terms is given by a constant number ‘d’ that we call the common difference. For example, the series 3, 7, 11, 15, 19, … is an Arithmetic Progression with 3 as the first term and 7 -3 = 15 -11 = 19 – 15 = 4 as the ‘d’ or the common difference.

If you know the first term or ‘a’, and the common difference ‘d’, you can construct the entire sequence. In fact the ‘nth’ term of an A.P. where ‘n’ is any number is equal to:  an = a + (n – 1)d; where ais the ‘nth’ term of the A.P. and ‘a’ is the first term. ‘d’ represents the common difference. this formula has many applications as follows:

Examples Based on A.P.

Example 1: How many numbers between 2 and 100 are divisible by 4?

Answer: Counting is not the thing you want to do here. In these types of questions, pick the lower number i.e. 2 and find the first number that is divisible by the number in the question ( 4 here). The first number that is divisible by 4 after 2 is 4 itself. So we have a starting point. Now each number divisible by 4 can be obtained from 4 by adding 4 or a multiple of 4 to our starting number. This will form an A.P. like 4, 8, 12, 16, … with a common ratio of 4 and the first term = 4.

Two Stage Type Series

But we don’t have to find any ‘nth tern’ rather we have to find the number of terms. This brings us to the second point which is to find the end number or the number just before 100 that is divisible by 4. The number is 100 itself. therefore our A.P. starts from 4 and ends at 100 and the number of terms or ‘n’ of this A.P. is the number of terms that are divisible by 4. Hence, from an = a + (n – 1)d we have: 100 = 4 + (n -1)4. This equation can be solved and the value of n that comes out is = 25. Therefore, there are 25 numbers between 4 and 100 that are exactly divisible by 4. let us move on to the two-stage type series now.

Two Stage Examples

Example 1: An A.P. has 2 as the first term and 3 as the common difference. A two stage type series that begins with 7 and contains this A.P. will have, ____ as its fifth term.

A) 18           B) 16             C) 24          D) 33

Answer: First we will form the A.P. and then use the terms got here to generate the two stage type series. The A.P. that has 2 as the first term and 3 as the common difference is 2, 5, 8, 11, 14, 17,20,… Now from this A.P. we shall form a two stage type series beginning with 7 i.e. 7, 7 + 2, 9 + 5, 14 + 8, 22 + 11, which is the fifth term. Thus 33 is the fifth term of the two stage type series. Thus the answer is D) 33.

Example 2: Find the missing term of the series: 7, 12, 20, 31, __

A) 39          B) 43         C) 63          D) 45

Answer: The series is clearly a two stage type series. We can easily verify it by finding the difference between the consecutive terms and making a table from them as follows:

5 8 11
7 12 20 31

The upper line represents an A.P. with a common difference of 3. The next term in this A.P. will be 14. Thus the next term in our two stage type series is = 31 + 14 = 45. Therefore the answer is D) 45.

Practice Questions

Q 1: How many two digit numbers are exactly divisible by 3?

A) 50           B) 33               C) 35          D) 30

Ans: D) 30.

Q 2: How many 3 digit numbers are exactly divisible by 4?

A) 300           B) 250           C) 225               D) 324

Ans: C) 225

Q 3: An A.P. has 6 as the first term and d = 2. The two stage type series that conatins this A.P. is:

A) 6, 12, 18, 24, 32

B) 2, 8, 24, 44, 64

C) 1, 7, 15, 25

D) 6, 13, 25, 37

Ans: C) 1, 7, 15, 25

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Jagdish
Guest

5 7 31 283 ?

Parth Joshi
Guest
Parth Joshi

4533

ajay jadhav
Guest
ajay jadhav

3967

Priya
Guest
Priya

5×1+2=7
7×4+3=31
31×9+4=283
283×16+5=4533

mahu
Guest

ook

Venu
Guest
Venu

4533

anmol
Guest
anmol

4,8,24,28,84,88,_

Parth Joshi
Guest
Parth Joshi

264

Rahul
Guest
Rahul

4+4=8×3=24+4=28×3=84+4=88×3=264

Ashish
Guest
Ashish

84+100=184or 84+240=324

spy7
Guest
spy7

4+4=8
8×3=24
24+4=28
28×3=84
84+4=88
88×3=264

pavan kalyan
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pavan kalyan

1 5 20 ???

lavanya lakhotia
Guest
lavanya lakhotia

60

anmol
Guest
anmol

1*3+2=5
5*3+5=20
20*3+7=67
can it not be like this???
if not then why?

Sathish
Guest

1×5=5
5×4=20
20×3=60

Saumit
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Saumit

How

srashti
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srashti

60

anmol
Guest
anmol

1*3+2=5
5*3+5=20
20*3+7=67
can it not be like this???
if not then why?

anmol
Guest
anmol

thats 20*3+8=68

Ashish
Guest
Ashish

1×2+3=5×3+5=20×4+7=86

Honey
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Honey

16,4,68,12,?,4,30,1,9 plz reply i need it

ABCD
Guest
ABCD

GAY

Ayushi shukla
Guest
Ayushi shukla

2,3,3,5,10,13,?,43,172,177

Sneha
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Sneha

4,5,5,7,9,13,10__,14

abcd
Guest
abcd

I think
Answer. 15
In series of odd numbers (4, 5,9,10,14) there is addition of 1 and 4 alternately.
And in series of even numbers (5, 7,13,?) There is addition of 2 and 6 alternately.

Janu
Guest
Janu

Find missing teams-1,5,14,?,44

Robin rathi
Guest
Robin rathi

Find the missing number of this series
60,50,60,90,41,_?

Options 1. 12
2. 18
3. 25
4. 30
5. none this above

Rohith
Guest
Rohith

26,4,20,10,14,16,8,22,2,28

Paulo
Guest

____,360,000,000,____, 389,000,000,____, 420,000,000

Sune Pedersen
Guest
Sune Pedersen

37 52 93 75 29 ? what is the math behind this

Sonali sahu
Guest
Sonali sahu

2,1,0,-3,-24,? Find the next number

Pinky Choudhary
Guest

Upper line 3 5 8 mid line 6 10 32 lower line 9 ? 50 me missing no. Kya h

Akash
Guest
Akash

94 101 115 136 164 ?

Ashish
Guest
Ashish

199

Lokesh
Guest
Lokesh

QID : 426 – In the following question, select the
missing number from the given alternatives.
41, 83, 167, 335, 671, ?
Options:
1) 1297
2) 1343
3) 1447
4) 1661

Ashish
Guest
Ashish

1343

Shriya Nimje
Guest
Shriya Nimje

50,50,54,72,?,220

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