Although the questions on progressions may not come directly in MBA exams, the theory behind progressions is used in every place where we need to sum up numbers. During your CAT 2007 or 2008 preparations, the formula used in progressions should become second nature to you as they will save a lot of time. Also, the methods of summing up various types of progressions, arithmetic, geometric or otherwise, should be very clear to you so that you are able to instantly spot the type of series you are facing. Knowing the basic methods of progressions also helps you simplify a lot of complex series. So let’s start with some basic progressions and their properties:

**Arithmetic Progression**

Numbers are said to be in Arithmetic Progression (A.P.) when the difference between any two consecutive numbers in the progression is constant i.e. in an Arithmetic Progression the numbers increase or decrease by a constant difference.

Each of the following series forms an Arithmetical Progression:

2, 6, 10, 14…

10, 7, 4, 1, -2…

a, a + d, a + 2d, a + 3d…

**Example:1. **If the 7

^{th}term of an Arithmetical Progression is 23 and 12

^{th}term is 38 find the first term and the common difference.

Answer: 7

^{th}term = 23 = a + 6d ---- (1)

12

^{th}term = 38 = a + 11d ---- (2)

Solving (1) and (2) we get a = 5 and d = 3

**2. **How many numbers of the series -9, -6, -3 … should we take so that their sum is equal to 66?

Answer: n[-18 + (n - 1)3]/ 2 = 66

n^{2} – 7n – 44 = 0

.·. n = 11 or -4.

The series is -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21..

We can see that the sum of first 7 terms is 0. The sum of next four terms after 7^{th} terms gives us the sum. Otherwise, if we count 4 terms *backward* from -9 we’ll get the sum as -66.

**3. **What is the value of k such that k + 1, 3k – 1, 4k + 1 are in AP?

Answer: If the terms are in AP the difference between two consecutive terms will be the same. Hence,

(3k – 1) – (k + 1) = (4k + 1) – (3k – 1)

2k – 2 = k + 2 --> k = 4.

**To insert arithmetic means between two numbers**

Let n arithmetic means m_{1}, m_{2}, m_{3}… m_{n} be inserted between two numbers a and b. Therefore, a, m_{1}, m_{2}, m_{3}, ….m_{n}, b are in arithmetic progression.

Let d be the common difference.

Since b is the (n + 2)^{th} term in the progression, b = a + (n + 1)d

Whence d = (b – a)/(n + 1)

Hence m_{1} = a + (b – a)/(n + 1), m_{2} = a + 2(b – a)/(n + 1).. and so on.

**Example:4. **If 10 arithmetic means are inserted between 4 and 37, find their sum.

**First Method:**

Let the means be m

_{1}, m

_{2}, m

_{3}… m

_{10}. Therefore 4, m

_{1}, m

_{2}, m

_{3}… m

_{10}, 37 are in AP and 37 is the 12

^{th}term in the arithmetic progression. Hence, 37 = 4 + 11d --> d = 3

Therefore means are 7, 10, 13 … 34 and their sum is 205.

**Second Method:**

We know that in an AP-

the sum of first term + last term = sum of second term + second last term = the sum of third term + third last term = .. and so on.

Therefore, 4 + 37 = m

_{1}+ m

_{10}= m

_{2}+ m

_{9}= m

_{3}+ m

_{8}= m

_{4}+ m

_{7}= m

_{5}+ m

_{6}= 41.

Hence m

_{1}+ m

_{2}+ m

_{3}… + m

_{9}+ m

_{10}= (m

_{1}+ m

_{10}) + (m

_{2}+ m

_{9})…+ (m

_{5}+ m

_{6})

= 5 x 41 = 205.

** **

**Example:5. **The sum of three numbers in A.P. is 30, and the sum of their squares is 318. Find the numbers.

Answer: Let the numbers be a – d, a, a + d

Hence a – d + a + a + d = 30 or a = 10

The numbers are 10 – d, 10, 10 + d

Therefore, (10 – d)

^{2}+ 10

^{2}+ (10 + d)

^{2}= 318

Or d = 3, therefore the numbers are 7, 10, and 13.

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