- Home
- Allegations and Mixtures
- AP GP HP
- Arithmetic Progressions
- Averages
- Boats and Streams
- Geometric Progressions
- Harmonic Progressions
- Clocks
- Calendar
- Clocks and Calendars
- Compound Interest
- Simple Interest
- Simple Interest and Compound Interest
- Linear Equations
- Quadratic Equations
- Co-ordinate geometry
- Perimeter Area Volume
- Divisibility
- HCF and LCM
- HCF
- LCM
- Number System
- Percentages
- Permutations Combinations
- Combinations
- Piipes and Cisterns
- Probability
- Work and Time
- Succesive Discounts
- Heights and Distance
- Decimals and Fractions
- Logarithm
- Venn Diagrams
- Geometry
- Set Theory
- Problem on Ages
- Inverse
- Surds and Indices
- Profit and Loss
- Speed, Time and Distance
- Algebra
- Ratio & Proportion
- Number, Decimals and Fractions

# How To Solve Arithmetic Progression Questions Quickly

## How To Solve Arithmetic Progression Questions Quickly

**Arithmetic progression is a sequence of terms such that the difference between the consecutive terms is a constant value . Or we can say that a succession of terms or numbers formed and arranged in a definite order according to a definite rule is called a Arithmetic Progression.**

**How to solve Arithmetic Progressions questions easily**

**Definition**

**Arithmetic progression (AP)** is a sequence of numbers in which each term after the first is obtained by adding a constant d to the preceding term. The constant d is called common difference.

An AP is represented in the form a, (a + d), (a + 2d), (a + 3d), …

where a = the first term, and d = the common difference.

General form, **T _{n} = a+ (n-1)d**

**Type 1: ****Find ****n**^{th }term of an AP: t_{n} = a + (n – 1)d

^{th }term of an AP:

**Question 1. If t _{n} = 20, d = -1 and n = 16, then the first term is?**

**Options:**

**A. 36**

**B.-35**

**C. 35**

**D. 5**

**Solution: ** We know that, t_{n} = a + (n – 1)d

20 = a + (16 – 1) * (-1)

20 = a + (15) * (-1)

20 = a – 15

-a = – 15 – 20

– a = – 35

a = 35

**Correct option: C**

**Question 2. What is the position of 62 in the following series 2, 5, 8, ….?**

**Options:**

**A. 21**

**B. 22**

**C. 23**

**D. 20**

**Solution: **In the given series,

a = 2

d = 3

t_{n} = 62

t_{n} = a + (n – 1)d

62 = 2 + (n -1) * 3

62 = 2 + 3n – 3

62 – 2 + 3 = 3n

63 = 3n

n = 21

**Correct option: A**

**Question 3. Find the 11 ^{th} term of the series 1, 3.5, 6, 8.5, ……**

**Options:**

**A. 25**

**B. 26**

**C. 23.5**

**D. 24.5**

**Solution: **In the given series a = 1, d = 2.5

n =11

t_{n} = a + (n – 1)d

t_{n} = 1 + (11 – 1) * 2.5

t_{n} = 1 + 10 * 2.5

t_{n} = 1+ 25

t_{n} = 26

**Correct option: B**

**Type 2: Number of terms in the series: n = \left [ \frac{l – a}{d} \right ] + 1**

**Question 1.Find the number of terms in the series 2, 4, 6, . . .98**

**Options:**

**A. 50**

**B. 49**

**C.51**

**D. 48**

**Solution **We know that, n = \left [ \frac{l – a}{d} \right ] + 1

a (first term) = 2

l (last term) = 98

d (common difference) = 2

n= [(98-2) / 2 ]+ 1

n = (96/2) + 1

n = 48 + 1

n = 49

**Correct option: B**

**Question 2.Find the last number of the series where, a = 3, d = 5, and number of terms are 45**

**Options:**

**A. 225**

**B.223**

**C.230**

**D. 222**

**Solution **According to the question,

a = 3, d = 5, n = 45

n = \left [ \frac{l – a}{d} \right ] + 1

45 = [(l-3)/5]+ 1

45-1 = (l-3)/5

44*5 = l – 3

220 = l-3

l=223

**Correct option: B**

**Question 3.Find the total odd numbers in the series 3….. 99.**

**Options:**

**A. 49**

**B.51**

**C. 50**

**D. 60**

**Solution: **The common difference is 2 because it is an odd number series.

We know that,

n = \left [ \frac{l – a}{d} \right ] + 1

a = 3

l = 99

d (common difference) = 2

n= (99-3)/2 + 1

n = 96/2 + 1

n = 48 + 1

n = 49

**Correct option: A**

**Type 3: Sum of first ‘n’ terms of the series: S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ] \text{ or } \frac{n}{2}(a + l)**

** **

**Question 1.Sanjana has one paisa on 1 ^{st} of January with her. She got 2 paisa on the 2^{nd} January and 3 paisa on 3^{rd} of January. In similar fashion, what will be her saving after 365 days?**

**Options:**

**A. 66700 paisa**

**B. 66795 paisa**

**C. 65795 paisa**

**D. 66790 paisa**

**Solution: **According to the question, the series formed is 1,2,3,4,5,……

Therefore, a = 1 and d = 1

S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ]

n = 365

S_{365 }= 365/2*[2 * 1 + (365 − 1) * 1]

S_{365 }= 66795 paisa

**Correct option: C**

**Question 2.Find the sum of the series 62 + 60 + 58 …….. + 30**

**Options:**

**A. 728**

**B.780**

**C.719**

**D. 782**

**Solution: **In the given series a = 62, d = -2, t_{n} = 30

t_{n} = a + (n – 1)d

30= 62 + (n – 1) * -2

30 = 62 – 2n + 2

2n = 62-30+2

34 = 2n

n = 17

S_{n} = \frac{n}{2}(a + l)

S_{17 }= 17/2 (62 + 30)

S_{17 }= 17/2(92)

S_{17} = 782

**Correct option: D**

**Question 3.If the sum of 5 consecutive numbers is 100 then find out the first number.**

**Options:**

**A. 20**

**B. 25**

**C. 18**

**D. 16**

**Solution: **It is clear from the question that d = 1 because there are 5 consecutive numbers

n = 5

S_{n}= 100

Let the first number ‘a’ be x

S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ]

100 = 5/2 [2a + (5 − 1) * 1]

100 = 5/2[2a + (4) * 1]

100*2/5 = [2a + 4]

40 = 2a + 4

a = 18

So, the first number is 18 and the complete series is 18, 19, 20, 21, and 22.

**Correct option: C**

**Type 4: Arithmetic mean of the series: **b = \frac{(a + c)}{2}

**Question 1.Find the arithmetic mean of first five prime numbers.**

**Options:**

**A. 4.5**

**B. 6.5**

**C. 5.6**

**D. 6.4**

**Solution: **We know that, arithemetic mean of a,b,c is

b** = ** (a + c)/2

Here, five prime numbers are 2, 3, 5, 7 and 11

Therefore, there arithmetic mean (AM) =(2+3+5+7+11)/5 = 5.6

**Correct option: C**

**Question 2. Find the arithmetic mean of first 11 whole numbers.**

**Options:**

**A. 4**

**B. 6**

**C.5**

**D.6**

**Solution: **We know that,

b = \frac{1}{2(a + c)}

Therefore, arithmetic mean (AM) = 0+1+2+3+4+5+6+7+8+9+10)/11= 5

**Correct option: C**

**Question 3. Find the arithmetic mean of series 2, 4, 6 , 8….100**

**Options:**

**A. 51**

**B.50**

**C.25**

**D.49**

**Solution: **In the given series, a = 2, d = 2, l = 100

To find the total number of term we can use the formula

n= [\frac{(l-a)}{d}] + 1

n = [(100-2)/2]+ 1

n = 50

Now, to find the sum of the series we can use the formula S_{n} = [2a + (n − 1) d] OR n/2 (a+l)

S_{n }= n/2 (a+l)

S_{n }= n/2(a+l)

S_{50} = 50/2(2+l00)

S_{50} = 25 (102)

S_{50 }= 2550

Now, AM = 2550/50 = 51

**Correct option: A**

**Read Also** –**Tips and Tricks to solve Boat and stream Problems **

Login/Signup to comment