## What is Arithmetic Progression?

This is a series of numbers where differences between any two consecutive number remain same.

**For example**: 2, 4, 6, 8 will be in AP since difference here is always 2

## Give some example of AP series.

Case | Example | Difference |

Consecutive Number | 1, 2, 3, 4, … 7, 8, 9, 10, … | 1 |

Even Numbers | 2, 4, 6, 8, … 90, 92, 94, … | 2 |

Odd Numbers | 1, 3, 5, 7, … 567, 569, 571, … | 2 |

Multiple of numbers | 11, 22, 33, 44, … 19, 38, 57, 76, … multiple of n | 11 19 n |

## Average of AP Series

If you are asked to find average of 1, 3, 5, you will find it as 3 which is middle number of given series. And the average is at 2^{nd} position.

Or if you are asked to find average of 1, 3, 5, 7, you will find it as 4 which is exactly middle number of given series. And the average is at at 2.5^{th} position

So, we could say that **In a AP series average lies exactly in the middle**.

We know, Sum of AP series = \displaystyle \frac{n}{2} \left(a+l\right)

Here, a = first term and l = last term

∴ Avg = \displaystyle\frac{a+l}{2}

Considering all,

Avg = Middle term of AP Series

= \displaystyle\frac{\text{first term}+\text{last term}}{2}

= \displaystyle\frac{\text{2nd term}+\text{2nd last term}}{2}

= ⋮

## How to find position of middle number in a AP series?

Odd number series with n element:

Middle term at = \displaystyle \left( \frac{n+1}{2}\right)^{\text{th}} position

Even number series with n element:

Middle term lies between \displaystyle \left( \frac{n}{2}\right) and \displaystyle \left( \frac{n}{2}+1\right)^{\text{th}} position

Here middle term at = \displaystyle \left( \frac{n}{2}+0.5\right)^{\text{th}}

**So, for both the cases, middle term at = \displaystyle \left( \frac{n+1}{2}\right)^{\text{th}} position**

## Average of squares and cubes of numbers in AP series

Average will be shifted to slight right in case of squares of numbers in AP series and more to right in case of cubes of numbers in AP series.

See the position of averages for 3 numbers and 4 numbers in the following picture.

## Problems on Average of AP Series

**Question**: The average of 7 consecutive numbers is 33. The largest of these numbers is

(a) 30

(b) 32

(c) 36

(d) 40

**Ans**: (c) 36

**Explanation**:

Middle term = \displaystyle\frac{7+1}{2} = 4^{th} = 33

Largest term (7^{th}) = 33 +3 = 36

**Question**: Average of 27 consecutive even numbers is 82. Find the largest and smallest number.

(a) 108, 56

(b) 109, 55

(c) 110, 54

(d) None of these

**Ans**: (a) 108, 56

**Explanation**:

It s a series of even numbers, hence gap between numbers = 2

Middle term = \displaystyle\frac{27+1}{2} = 14^{th} = 82

Largest term (27^{th})

= 82 + 2(27 – 14)

= 82 + 26

= 108

Smallest term (1^{st})

= 82 – 2(14 – 1)

= 82 – 26

= 56

**Aliter(By Formula)**

This formulae are only applicable to consecutive even and odd numbers.

**Highest No = avg + (n – 1)Lowest No = avg – (n – 1)**

Here, avg = 82

and n – 1 = 27 -1 = 26

∴ Highest No = 82 + 26 = 108

Lowest score = 82 – 26 = 56

**Question**: Average of 14 consecutive odd numbers is 38. Find the largest and smallest number.

(a) 50, 26

(b) 51, 25

(c) 52, 24

(d) None of these

**Ans**: (b) 51, 25

**Explanation**:

It s a series of odd numbers, hence gap between numbers = 2

Middle term = \displaystyle\frac{14+1}{2} = 7.5^{th} = 35

Largest term(14^{th})

= 38 + 2(14-7.5)

= 38 +13

= 51

Smallest term (1^{st})

= 38 – 2(7.5-1)

= 38 – 13

= 25

**Aliter(By Formula)**

Highest No = 38 + 13 = 51

Lowest score = 38 – 13 = 25

[**Similar Question**] ⟶ 2017 (Q2) of Average PYQs of CGL

**Question**: Average of 13 consecutive multiples of 11 is 14641. Find the second largest number of the series

(a) 14652

(b) 16496

(c) 14696

(d) 14608

**Ans**: (c) 14696

**Explanation**:

Here 1 gap = 11

Middle term = \displaystyle\frac{13+1}{2} = 7th = 14641

Second largest term (12^{th})

= 14641 + 11(12-7)

= 14641 + 11 × 5

= 14641 + 55

= 14696

**Question**: Average of 8 consecutive multiples of 17 is 49189.5. find the second smallest number of the series.

(a) 49197

(b) 49497

(c) 49147

(d) 49749

**Ans**: (c) 49147

**Explanation**:

Here, 1 gap = 17

Middle term = 4.5^{th} = 49189.5

Second smallest no (2^{nd})

= 49189.5 – 17(4.5 – 2)

= 49189 – 17 × 2.5

= 49189 – 42.5

= 49197

**Question**: Average of 6 consecutive multiples of 19 is 4322.5. find the second largest number of the series.

(a) 4313

(b) 4332

(c) 4351

(d) 4370

**Ans**: (c) 4351

**Explanation**:

Here, 1 gap = 19

Middle term = 3.5^{th} = 4233.5

Second largest (5^{th})

= 4322.5 + 19(5 – 3.5)

= 4322.5 + 19 × 1.5

= 4322.5 + 28.5

= 4351

**Question**: The average of squares of five consecutive odd natural numbers is 233. What is the average of the largest and smallest number?

(a) 15

(b) 17

(c) 11

(d) 13

**Ans**: (a) 15

**Explanation**:

Middle term = 3rd

Average of squares lies between 3rd and 4th

Now a square of odd number which is immediate less than 233 is 225 which is 15^{2}

So, we have got 3rd term = 15 and got the other term accordingly, since it’s a odd number series, gap = 2

Avg = middle term = \displaystyle\frac{\text{first term}+\text{last term}}{2}

Hence, avg = 15

**Aliter(By Traditional Method)**

Let, the numbers are a-4, a-2, a, a+2 and a+4

According to the question,

(a-4)^{2} + (a-2)^{2} + a^{2} + (a+2)^{2} + (a+4)^{2} = 233 × 5

⇒ {(a-4)^{2} + (a+4)^{2}} + {(a-2)^{2} + (a+2)^{2}} + a^{2} = 233 × 5

⇒ 2(a^{2} + 16) + 2(a^{2} + 4) + a^{2} = 233 × 5

⇒ 5a^{2} + 40 = 233 × 5

⇒ a^{2} + 8 = 233 [Dividing both side by 5]

⇒ a^{2} = 225

⇒ a = 15

Hence, numbers are 11, 13, 15, 17, 19

Traditional method will consume more time. So we will go for the previous technique.

**Question**: The average of squares of four consecutive odd natural numbers is 201. The average of 7 times the largest and 3 times the smallest number is :

(a) 72

(b) 78

(c) 76

(d) 66

**Ans**: (c) 76

**Explanation**:

**Question**: The average of squares of four consecutive even numbers is 126. The average of 8 times the greatest and 5 times the smallest number is

(a) 66

(b) 76

(c) 68

(d) 74

**Ans**: (b) 76

**Explanation**:

**Question**: The sum of 5 consecutive odd numbers is 215. What is the sum of a different set of 5 consecutive numbers whose second lowest number is 37 less than double of the lowest of the first set?

(a) 210

(b) 230

(c) 250

(d) 240

**Ans**: (a) 210

**Explanation**:

Hence, sum of second series = 42 × 5 = 210

You may find the all numbers of S1 and S2, but we have found only those required for calculation.

**Question**: S1 is the series of five consecutive multiples of four, whose sum is 100 and S2 is the series of four consecutive even numbers. If the second number of S2 series is 6 less than highest number of S1 series, find the average of S2 series.

(a) 32

(b) 23

(c) 25

(d) 21

**Ans**: (b) 23

**Explanation**:

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