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As the Tutorial Focus, mastering this mathematical aspect becomes pivotal in enhancing problem-solving abilities and ensuring confidence during the JAMB examination.

Hi guys. Today we’ll be looking to completely exhaust **sequences and series** based on **jamb mathematics syllabus**.

If you feel you need a good textbook instead, **see the recommended textbook jamb asked students to read for mathematics**.

However, for those who feel they have completely exhausted this topic, **click here to see tough questions on sequences and series**

Having said all that, let’s proceed.

**For the best guide in your studies and mathematics tutorials with videos, visit ogschools.com**

## 1.1 Sequences and series

A set of numbers such as $U_{1},U_{2},U_{3},U_{4},…,U_{n}$ which are ordered is called a sequence. $U_{1}$ is the first term, $U_{2}$ is the second term, $U_{n}$ is the $n_{th}$ term of the sequence. The following are examples of sequences

**1. **$1,3,5,7,…,2n−1$ (sequence of odd numbers)

**2. **$2,3,6,8,10,…,2n$ (sequence of even numbers)

The expression $U_{1}+U_{2}+U_{3}+U_{4}+…+U_{n}$ which can be formed from a sequence is called a series.

## 1.2 Arithmetic progression (A.P)

An arithmetic progression of A.P is a series of the form: $a+(a+d)+(a+2d)+(a+3d)+…$ where a is the first term and d is called the common difference which may be positive or negative. Each bracket of the A.P is a term and simple progression will give the $n_{th}$ term of the sequence.

$U_{n}=a+(n−1)d$

If there are n terms in the A.P and $S_{n}$ is their sum, then

$S_{n}=[a]+[a+d]+[a+2d]+…+[a+(n−1)d]$

Reversing the series gives

$S_{n}=[a+(n−1)d]+[a+(n−2)d]+[a+(n−3)d]+…+[a]$

Adding both equations of $S_{n}$gives

$2S_{n}=[2a+(n−1)d]+[2a+(n−1)d]+[2a+(n−1)d]+…+[2a+(n−1)d]2S_{n}=n[2a+(n−1)d]S_{n}=2n [2a+(n−1)d]${Sum of n terms of a sequence}

$U_{1}=$ first term $=a,U_{n}=$ last term $=a+(n−1)d∴S_{n}=2n [U_{1}+U_{n}]$ {Sum of n terms of a sequence}

## 1.3 Arithmetic mean

If $p,q,r$ are three consecutive terms of an A.P, then q is called the arithmetic mean of p and r. Since $p,q,r$ are consecutive terms of an A.P, it implies that

$q−p=r−q=dq−p=r−q2q=r+pq=2r+p $

The arithmetic mean of two numbers is the average of the two numbers.

**Example 1**

**a.** Find the $1_{th}$ term and the sum of the first 26 terms of the series $21,17,13,9,5,…$

**b. **Calculate the arithmetic mean of -15 and -23

**Solution**

Given $21,17,13,9,5,…a=21,d=17−21=−4,n=15,U_{15}=?,S_{26}=?$

**a. i. **$U_{n}=a+(n−1)dU_{15}=21+(15−1)(−4)U_{15}=21+(14)(−4)U_{15}=21−56U_{15}=−35$

**ii. **$S_{n}=2n [2a+(n−1)d]S_{26}=226 [2(21)+(26−1)(−4)]S_{26}=13[42+(25)(−4)]S_{26}=13[42−100]S_{26}=13[−58]S_{26}=−754$

alternatively,

$U_{26}=21+25(−4)U_{26}=21−100U_{26}=−79S_{n}=2n [U_{1}+U_{n}]S_{26}=226 [21+(−79)]S_{26}=13[−58]S_{26}=−754$

**b.** arithmetic mean $=2+() =−19$

## 1.4 Geometric progression (G.P)

A series that is in the form $a+ar+ar_{2}+ar_{3}+…$ is called a geometric progression, where a is the first term and r is the common ratio. The $n_{(}th)$ term is $U_{n}=ar_{n−}$.

Let $S_{n}$ be the sum of the $n_{th}$ terms of a series

$S_{n}=a+ar+ar_{2}+ar_{3}+…+ar_{n−}$

multiply through by r

$rS_{n}=ar+ar_{2}+ar_{3}+…+ar_{n−}+ar_{n}$

subtracting $rS_{n}$ from $S_{n}[/latex] gives

$(1−r)S−n=a−ar_{n}S_{n}=−ra(−r) $ (first equation)

subtracting $S_{n}$ from $rS_{n}$ gives

$(r−1)S_{n}=ar_{n}−aS_{n}=r−a(r−) $ (second equation)

Both equations can be used to solve for $S_{n}$

## 1.5 Geometric mean

If the number $a,b,c$ are three consecutive terms of a G.P, then b is called the geometric mean of a and c. Since $a,b,c$ are consecutive terms of a G.P,

$ab =bc =rb_{2}=acb=ac $

The value of b could either be +ve or -ve but never both. The common ratio determines it.

**Example 2**

**a. **Find the sum of the first 8 terms of the series $271 ,91 ,31 ,…$

**b. **If the $_{th}$ and $1_{th}$ terms of a G.P are 320 and 2560 respectively, find the series

**Solution**

**a. **Given $271 ,91 ,31 ,…a=271 ,r=91 ÷271 =3,n=8S_{n}=−ra(−r) S_{8}=−271 (−) =×−−6560 S_{8}=−54−6560 S_{8}=121.48$

alternatively,

$S_{n}=r−a(r−) S_{8}=−271 (−) =×6560 S_{8}=546560 S_{8}=121.48$

It’s obvious that both equations can work

**b. **$_{th}$ term $=ar_{−}=ar_{6}ar_{6}=320$

Similarly, $1_{th}$ term $=ar_{9}ar_{9}=2560arar =3202560 r_{3}=8r=38 r=2ar_{6}=320a(2_{6}=320a=64320 a=5$

Remember a geometric series is represented as $a,ar,ar_{2},ar_{3},…∴$ the G.P is $5,10,20,40$

### 1.6 Sum to infinity of a G.P

$S_{∞}=−ra $

Where $/r/<1$ i.e the magnitude of r whether positive or negative is always less than 1.

$∴$ ranges from -1 to + 1 i.e $−1<r<1$ but $r0$

Thanks for taking out time to read this article. **If there’s any area you feel was not touched, needs correction or needs clarification, do well to use the comment section below**.

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