**Series and progression.**

In the quantitative aptitude section of CAT exam, “series and progression” is considered as one of the most important topic. Series and progression are basically the itemized collection of numbers which are arranged in a certain predictable order.**Arithmetic Progression**

The mathematical arrangement of the numbers where the difference between any the consecutive numbers is always equal (Constant) is generally termed as Arithmetic Progression or AP (in Short).

Example: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28. (AP of 10 terms with difference = 3)

Example: 14, -2, -18, -34, -50, -66 … … … ∞ (Infinite AP with difference = -16)

Inference 1 – In AP every next term can be obtained by adding the common difference in the previous term.

Inference 2 – Only the first term and common difference is required to define an Arithmetic Progression up to infinity.**Terminologies of Arithmetic Progression**

\(\Rightarrow T_1 =\) First term of AP

\(\Rightarrow T_2 =\) Second term of AP

\(\Rightarrow T_3 =\) Third term of AP ………so on If

n = total no. of terms in AP

∴ \(T_n = \) Last term of AP

d = common difference of AP (\(d = T_n – T_{n-1}\))

\(S_n\) = Sum of AP**General form of Arithmetic Progression**

\(\Rightarrow\) \(T_1, T_2, T_3, \dots\dots \dots , T_{n-1}, T_n\) \(\dots\dots \dots (1)\)

Where,

Let \(T_1 = a\) , So as per the Inference 1

\(T_2 = a + d\)

\(T_3 = a + 2d\)

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\(T_{14} = a + 13d\)

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\(T_n = a + (n-1)d\)

Therefore equation (1) can also be written as

\(\Rightarrow\) \(a, a+d, a+2d, \dots\dots \dots , a+(n-2)d, a+(n-1)d\) \(\dots\dots \dots (2)\)

**(Q.1) Find the \(20^{th}\) term of the following AP1, 4, 7, 10, 13, 16 … … …**

Solution:

Here, \(a = 1,\) \(d = (T_2 – T_1) = 4 -1 = 3\)

\(T_{20} = ?\)

we know that \(n^{th}\) term of AP is given by:

\(\Rightarrow\) \(T_n = a + (n-1)d\)

By putting the value of a, d and n,

\(\Rightarrow\) \(T_{20} = 1 + (20-1)3\)

\(\Rightarrow\) \(T_{20} = 1 + (19)3\)

\(\Rightarrow\) \(T_{20} = 58\)

**(Q.2) Find the number of terms in the following AP5, 9, 13, 17, 21, 25, … … … , 401.**

Solution:

\(T_n = 401,\) \(a = 5,\) \(d = 4\)

we know that \(n^{th}\) term of AP is given by:

\(\Rightarrow\) \(T_n = a + (n-1)d\)

By putting all the values,

\(\Rightarrow\) \(401 = 5 + (n-1)4\)

\(\Rightarrow\) \(396 = 4n – 4\)

\(\Rightarrow\) \(4n = 400\)

\(\Rightarrow\) \(n = 100)\)

Hence, there are 100 terms in the given AP.

**Graphical Representation**

AP with common difference 2 Versus GP with common ratio 2

AP follows linear values where as GP follows exponential values

Let’s consider an AP: **-3, 3, 9, 15, 21, 27, 33**.

Average of the AP = \(\frac{\text{Sum of all terms}}{\text{Total no. of terms}} \dots\dots \dots \) (3)

Average of the AP = \(\frac{105}{7}= 15\)

So, we can say that the middle term of the AP is equal to its average

Or, the average of \(n^{th}\) term from first and the last in the AP is equal to the average of whole AP

Let \(a\) be the first term and so the last term will be \(a+(n-1)d\)

\(\Rightarrow\) Avg. of AP = \(\frac {\text{First term + Last term}}{2}\)

\(\Rightarrow\) \(\frac{ \text{Sum of all terms}}{\text{Total no. of terms}}\) = \(\frac {\text{First term + Last term}}{2}\)

\(\Rightarrow\) \(\frac{S_n}{n} = \frac{a + a+(n-1)d}{2}\)

\(\Rightarrow\) \(S_n = \frac{n}{2}\times (2a+(n-1)d\)

**(Q.3) Find the sum of the given AP up to 21 terms-2, 3, 8, 13, … … … (21 terms)**

Solution:

We have to find the sum of 21 terms (i.e. \(S_{21}\)) ∴ \(n = 21\), \(a = -2\) and \(d = 5\)

Sum of first n terms of AP is given by \(\Rightarrow S_n = \frac{n}{2}\times (2a+(n-1)d)\)

\(\Rightarrow\) \(S_{21} = \frac{21}{2}\times (2(-2)+(21-1)5)\)

\(\Rightarrow\) \(S_{21} = \frac{21}{2}\times (96)\)

\(\Rightarrow\) \(S_{21} =1008\)

**(Q.4) If sum of first 20 terms of AP is 1050, whose first term is 100. Then find the last term**

Solution:

Given that \(S_{20} = \frac{20}{2}\times (2(100) + (20-1)d\)

\(\Rightarrow\) \(1050 = 10 \times (200 + 19d)\)

\(\Rightarrow\) \(d = -5\)

Last term i.e. \(T_{20} = 100+(19)(-5) = 5\)

**Selection of terms in an AP**

Sometimes we have to assume certain numbers of terms in AP for the ease of calculation. The following ways are usually used for the selection of terms in an AP.

Selection of 3 terms in AP: Assume these 3 numbers as – \(a-d,\) \(a,\) and \(a+d\)

Selection of 4 terms in AP: Assume these 4 numbers as – \(a-3d,\) \(a-d,\) \(a+d,\) and \(a+3d\)

Selection of 5 terms in AP: Assume these 5 numbers as – \(a-2d,\) \(a-d,\) \(a,\) \(a+d,\) and \(a+2d\)

**(Q.5) The sum of five terms of an A.P. is 200 then find the middle term.**Solution:

Let’s consider the three terms of the A.P. to be \(a-2d,\) \(a-d,\) \(a\) \(a+d\) and \(a+2d\)

\(\Rightarrow\) \(a-2d+a-d+a+a+d+a+2d=200\)

\(\Rightarrow\) \(5a= 200\)

\(\Rightarrow\) \(a=40\)

Hence the middle term is equal to 40.

**(Q.6) The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.**

Solution:

Let’s consider the three terms of the A.P. to be \(a-d, a, a+d\)

so, the sum of three terms \(=21\)

\(\Rightarrow a-d+a+a+d=21\)

\(\Rightarrow 3 a=21\)

\(\Rightarrow a=7\)

And, product of the \(1^{st}\) term and \(3^{rd}\) term =\( 2^{nd}\) term + 6

\(\Rightarrow\) \((a-d)(a+d)=a+6\)

\(a^{2}-d^{2}=a+6\)

\(\Rightarrow\) \((7)^{2}-d^{2}=7+6\)

\(\Rightarrow\) \( 49-d^{2}=13\)

\(\Rightarrow\) \( d^{2}=49-13=36\)

\(\Rightarrow\) \( d =\sqrt 6\)

\(\Rightarrow\) \( d=\pm6\)

Hence, the terms can be \(\Rightarrow 1,7,13\) or \(13,7,1\)

**Harmonic Progression**

A Harmonic Progression (HP) is defined as a sequence of real numbers which is formed by taking the reciprocals of the terms of an arithmetic progression that does not contain 0.

In harmonic progression, any term in the sequence is considered as the harmonic means of its two neighbours. For example, the sequence p, q, r, s, …is considered as an arithmetic progression; the harmonic progression can be written as \( \frac{1}{p}, \frac{1}{q}, \frac{1}{r},\frac{1}{s}, …\)

Example: \( \frac{1}{4}, \frac{1}{7}, \frac{1}{10},\frac{1}{13}, \frac{1}{16} …\)**Harmonic Mean**: Harmonic mean is calculated as the reciprocal of the arithmetic mean of the reciprocals. The formula to calculate the harmonic mean is given by:

Harmonic Mean = \(\frac{n}{\frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{1}{s} +….}\)

Where, p, q, r, s are the values and n is the number of values present.

**(Q.7) Compute the 16th term of HP if the 6th and 11th term of HP are 10 and 18, respectively.**

Solution:

The H.P is written in terms of A.P are given below:

6th term of \(\mathrm{A.P}=\mathrm{a}+5 \mathrm{~d}= (\frac{1}{10})–(\mathrm{I})\)

11th term of \(\mathrm{A} \cdot \mathrm{P}=\mathrm{a}+10 \mathrm{~d}=\frac{1}{18}\)

By solving these two equations, we get

\(a=\frac{13}{90}\) and \(d=\frac{-2}{225}\)

To find 16 th term, we can write the expression in the form.

\(a+15 d=\frac{13}{90}-\frac{2}{15} =\frac{1}{90}\)

Thus, the 16 th term of an (\mathrm{H.P}= \frac{1}{16}) th term of an (\mathrm{A.P}=90)

Therefore, the 16 th term of the H.P is 90.