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# How To Solve Arithmetic Progression Questions Quickly

## How to Solve Arithmetic Progression Questions in Aptitude

Arithmetic progression is a sequence of terms such that the difference between the consecutive 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.Although we know Arithmetic Progression but most of us don’t know How to Solve Arithmetic Progression Questions.This is very important to Know for us.

In this Page How to Solve Arithmetic Progression Questions is given along with their Solutions.This will very useful for students Preparing for different Examination.

**How to Solve Arithmetic Progression 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.

n is number of Terms

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

Where T_{n }is nth term of an Arithmetic Progresion_{ }

**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 x 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= [\frac{(98-2)}{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 = [\frac{(l-3)}{5}]+ 1

45-1 = \frac{(l-3)}{5}

44 x 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 = \frac{(99-3)}{2} + 1

n = \frac{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 }= \frac{365}{2}\times [2 \times 1 + (365 − 1) \times 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 }= \frac{17}{2}\times (62 + 30)

S_{17 }= \frac{17}{2}\times (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 = \frac{5}{2} [2a + (5 − 1) \times 1]

100 = \frac{5}{2}[2a + (4) \times 1]

100\times \frac{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 10 and 18.

Options:

A. 14

B. 15

C. 16

D. 17

**Solution: **We know that, Arithmetic mean of a and b is \frac{a+b}{2}

Therefore, there arithmetic mean (AM) = \frac{10+18}{2} = 14

Correct option: A

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

Options:

A. 4

B. 6

C.5

D.6

**Solution: **Here First term = 0

Last term = 10

Therefore, arithmetic mean (AM) = \frac{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 = [\frac{(100-2)}{2}]+ 1

n = 50

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

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

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

S_{50} = \frac{50}{2}(2+l00)

S_{50} = 25 (102)

S_{50 }= 2550

Now, AM = \frac{2550}{50} = 51

Correct option: A

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