Arithmetic Geometric Progression Part1

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Notes

For the same reason with quadratic (having learnt it in SPM), this topic could actually end up being more difficult than the rest of the subtopics in Sequences & Series. Pay attention to the concepts and not just the formulas (there aren't any new formula after all compared to SPM add maths.)

Learning Objectives (Syllabus)

use the general formula for the general term of an arithmetic progression
derive and use the formula for the sum of the first $n\,$ terms of an arithmetic series
solve problems involving arithmetic progressions and series
use the general formula for the general term of an geometric progression
derive and use the formula for the sum of the first $n\,$ terms of an geometric series
use the formula for the sum to infinity of a convergent geometric series
solve problems involving geometric progressions and series

Prior Knowledge

basic algebraic skills
index & logarithms (equality and also inequality)
a general ability to recognize patterns

Arithmetic Progression

Examples

$14,\,$ $17,\ldots$
- $2 \xrightarrow{{\color{Blue}+3}} 5 \xrightarrow{{\color{Blue}+3}} 8\xrightarrow{{\color{Blue}+3}} 11\xrightarrow{{\color{Blue}+3}}{\color{Red}14}\xrightarrow{{\color{Blue}+3}} {\color{Red}17}$
$-11,\,$ $-18,\ldots$
- $10 \xrightarrow{{\color{Blue}-7}} 3\xrightarrow{{\color{Blue}-7}} -4\xrightarrow{{\color{Blue}-7}}{\color{Red}-11}\xrightarrow{{\color{Blue}-7}} {\color{Red}-18}$
$x+3y,\,$ $x+4y,\ldots$
- $x \xrightarrow{{\color{Blue}+y}} x+y\xrightarrow{{\color{Blue}+y}} x+2y\xrightarrow{{\color{Blue}+y}}{\color{Red}x+3y}\xrightarrow{{\color{Blue}+y}} {\color{Red}x+y}$

Notation

First, differentiate between sequences and series

Sequences

$2,5,8,11\,$ - arithmetic progression

Series

$2+5+8+11\,$ - arithmetic series

An easy (but not terminologically correct) way to see it, is when we take the terms in the sequence and then

$u_{n}\,$ represents the $n^{th}\,$ term

$u_{1}\,$ represents first term
$u_{5}\,$ represents $5^{th}\,$ term
$u_{n+1}\,$ represents $(n+1)^{th}\,$ term
$u_{n-1}\,$ represents $(n-1)^{th}\,$ term

$S_{n}\,$ represents sum of the first $n\,$ terms (Sum of first term to $n^{th}\,$ term)

$S_{3}=\,$ $u_{1}+u_{2}+u_{3}\,$
$S_{10}=\,$ $u_{1}+u_{2}+u_{3}+\ldots+$ $u_{10}\,$
Hmmm... should we still start from ? Yes.
- $S_{n-1}=\,$ $u_{1}+u_{2}+u_{3}+\ldots+$ $u_{n-1}\,$

Hmmm... should we go ?
- $S_{2n}=\,$ $u_{1}+u_{2}+u_{3}+\ldots+$ $u_{2n}\,$

u_n for Arithmetic Progression

An arithmetic progression (AP from here onwards) takes the following form

$a, a+d, a+2d,\,$ $a+3d, a+4d,\ldots$
where
- $a=\,$ first term
- $d=\,$

Which is which term?

$\underbrace{a}_{1^{st} term},\underbrace{a+d}_{2^{nd} term},\underbrace{a+2d}_{3^{rd} term},\underbrace{a+3d}_{4^{th} term},\ldots$ Note the numbers $\underbrace{a+{\color{Blue}0}d}_{{\color{Blue}1}^{st} term},\underbrace{a+{\color{Blue}1}d}_{{\color{Blue}2}^{nd} term},\underbrace{a+{\color{Blue}2}d}_{{\color{Blue}3}^{rd} term},\underbrace{a+{\color{Blue}3}d}_{{\color{Blue}4}^{th} term},\ldots$

From the pattern, we can see that

$u_{7}=\,$ $a+6d\,$
$u_{50}=\,$ $a+49d\,$

And thus, the general formula for an arithmetic progression,

$a+{\color{Red}??}d\,$
- $u_{n}=a+\left(n-1\right)d$

Examples

$a+nd\,$
- Its because ${\color{Red}\left(n+1\right)}-1=n$
$u_{n-1}=\,$ $a+\left(n-2\right)d\,$
$a+\left(2n-1\right)d\,$
- Its simply $u_{{\color{Red}2n}}=a+\left({\color{Red}2n}-1\right)$

Definition of AP

As we can see above, the formula for AP does not define what an AP is, it is the other way round. We understand what an AP is, and then, we found a general formula to fit it. So how do we define exactly what is an AP?

$2,5,8,11,\ldots$ is an AP as $2\xrightarrow{+{\color{Blue}3}}5\xrightarrow{+{\color{Blue}3}}8\xrightarrow{+{\color{Blue}3}}11\ldots$
$2,5,9,12,\ldots$ is not AP as $2\xrightarrow{+{\color{Blue}3}}5\xrightarrow{+{\color{Red}4}}9\xrightarrow{+{\color{Blue}3}}12\ldots$
In other words, "something" must be always the same. That "something" is the difference.
To be more precise, we see that we will only take the differences of terms next to each other.
For that, we will use the term successive terms.
To say that it is always the same, we use the term constant.

Thus the full definition will be

Sequence where

Of course, we are more interested in an equation to represent the above definition, but let's try some examples first.

Note: It is WRONG to say

The common difference is constant.

Examples

Prove that the following forms arithmetic progressions

$3,7,11\,$
$u_{n}=4n-1\,$ . Find also the first term and common difference.

Let's do the first one

- Analyze :
- $u_{3}-u_{2}=11-7=4\,$
- $u_{2}-u_{1}=7-3=4\,$
- Note :

How about the second one?

- Analyze:
- $u_{n+1}-u_{n}=\,$ $\left[4\left(n+1\right)-1\right]-4n-1$
- $=4n+3-4n+1\,$
- $\mbox{Thus,}\;u_{n}\;\mbox{is an AP}$
  - Note : Writing $\begin{align} & u_{n+1}-u_{n}=d \\ & \ldots \\ & \therefore d=4\end{align}$ will end with zero marks
- $\mbox{First term} = u_{1}= 3\,$
- $\mbox{Common difference} = 4\,$

Definition(Formula)

Thus, a sequence will be an AP if $u_{n+1}-u_{n}=\mbox{constant}\,$

Use this definition when asked to prove a sequence is an AP.

Arithmetic Mean

If $a,b,c\,$ forms an AP, $b\,$ is said to be the arithmetic mean of $a\,$ & $c\,$

$\mbox{Since}\;a, b, c\;\mbox{forms an AP}$
$\therefore$
$\therefore b=$ $\frac{a+c}{2}$

Note : Arithmetic mean of ANY two values, $p\,$ & $q\,$ is $\frac{p+q}{2}$

In other words, arithmetic mean is just the "normal" mean/average.

S_n for Arithmetic Progression

Example

As the story goes, a smart kid figured how to find out this sum quickly.
- - Write out the sum. $S=1+2+3+\ldots +98+99+100$
  - Rewrite out the sum backwards. $S=100+99+98+\ldots +3+2+1$
  - Note that $1+100=101, 2+99=101, 3+98=101\,$ and so on...
  - $\begin{array}{ccccccccccccccc} S= & 1 & + & 2 & + & 3 & + & \ldots & + & 98 & + & 99 & + & 100 & \frac{\quad}{}(1)\\ S= & 100 & + & 99 & + & 98 & + & \ldots & + & 3 & + & 2 & + & 1 & \frac{\quad}{}(2)\\ \end{array}$
  - $(1)+(2):\,$
  - $\begin{array}{ccccccccccccccc} 2S= & 101 & + & 101 & + & 101 & + & \ldots & + & 101 & + & 101 & + & 101\\ \end{array}$
  - So how many $101\,$ 's are there?
  - $2S= \underbrace{101 + 101 + 101 + \ldots + 101 + 101 + 101}_{100\;terms}$
  - $2S= \left(101\right)\left(100\right)$
  - $S= \frac{\left(101\right)\left(100\right)}{2}=5050$

Prove

The above is a actually an arithmetic series $(a=1, d=1)\,$ , thus we can use the same method to derive a general formula for $S_{n}\,$ for any arithmetic series.

- First let's write out the sum. $a+\,$ $\left(a+d \right)+$ $\left(a+2d \right)+\ldots$
- The last term will be
- Note that since we need to "add to" the first 3 terms, we would also need to write out the
- $S_{n}\,$ $a+\left(a+d \right)+\left(a+2d \right)+\ldots$ $+\left(l-2d\right)+\left(l-d\right)+l$
- Note that when writing, we should leave enough between the first 3 terms to "fit" the last three terms
- Let rewrite the sum backwards, "pairing" the terms
  - $\begin{array}{ccccccccccccccc} S_{n}= & a & + & \left(a+d \right) & + & \left(a+2d \right) & + & \ldots & + & \left(l-2d \right) & + & \left(l-d \right) & + & l & \frac{\quad}{}(1)\\ S_{n}= & l & + & \left(l-d \right) & + & \left(l-2d \right) & + & \ldots & + & \left(a+2d \right) & + & \left(a+d \right) & + & a & \frac{\quad}{}(2)\\ \end{array}$
- And then we
- Thus $2S_{n}=\,$ $n\left(a+l\right)$
- $\therefore S_{n}=\frac{n}{2}\left(a+l\right)$
Thus we have
- But $l=\,$ $u_{n}\,=$ $a+\left(n-1\right)d$
- Thus, $S_{n}=\,$ $\frac{n}{2}\left\{a+{\color{Blue}\left[a+\left(n-1\right)d\right]}\right\}=$ $\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Formula

When to use which formula?

$S_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$ Use when we don't have last term.
Note : Don't mix up: $u_{n}={\color{Blue}a}+\left(n-1\right)d$ but $S_{n}=\frac{n}{2}\left[{\color{Red}2}{\color{Blue}a}+\left(n-1\right)d\right]$

Examples

Find the sum of the following series

$7+4+1+\ldots$ until the $20^{th}\,$ term
- Analyze :
- $\underbrace{7+4+1+\ldots u_{20}}_{\mbox{AP with}\;a=7,\; d=-3,\; n=20}$
- $=-430\,$

$2+5+8+\ldots+296$
- Analyze :
- $\underbrace{2+5+8+\ldots+296}_{\mbox{AP with}\;a=2,\; d=3,\; u_{n}=296}$
- $u_{n}=296\,$
- $2+\left(n-1\right)3=296$
- $\therefore n=99$
- $\therefore S_{99}=\frac{99}{2}\left(2+296\right)=14751$

Sum NOT from the first term

Sum from

term to term $=S_{8}-S_{4}\,$
- - The problem is $S_{n}\,$ always starts from the first term.
  - $S_{8}\,$ represents $u_{1}+u_{2}+u_{3}+\ldots +u_{8}$
  - But we only want $u_{5}+u_{6}+u_{7}+u_{8}\,$
  - $\overbrace{u_{1}+u_{2}+u_{3}+u_{4}+{\color{Blue}u_{5}+u_{6}+u_{7}+u_{8}}}^{S_{8}}$
  - Thus, we have to
  - Thus, what we need is $S_{8}-S_{4}\,$

Sum of the first 10 terms $=S_{10}\,$
Sum of the NEXT 10 terms $=S_{20}-S_{10}\,$
- The next 10 terms starts from $u_{{\color{Blue}11}}\,$ and ends with $u_{{\color{Blue}20}}\,$
  - Thus, it is $S_{{\color{Blue}20}}-\,$ $S_{{\color{Red}10}}\,$
  - NOT $S_{11}\,$ ${\color{Red}u_{1}+\ldots+u_{10}}+{\color{Blue}u_{11}+\ldots+u_{20}}$

Sum of the first n terms $=S_{n}\,$
Sum of the NEXT n terms $=S_{2n}-S_{n}\,$
- You can always write out the sums to guide you if you can't see it directly.
  - $\overbrace{\underbrace{{\color{Red}u_{1}+\ldots+u_{n}}}_{n\;terms\;/S_{n}}+\underbrace{{\color{Blue}u_{n+1}+\ldots+u_{2n}}}_{n\;terms}}^{S_{2n}}$

Interpreting Information

The first step to interpret the given information into equations is very crucial as any mistake will almost surely result in zero marks (and lots of wasted time). Read carefully and do not rush, especially in exams.

term is and sum of first terms is
- $u_{5}=6,\;S_{5}=50\,$
The difference between the and term is times the term
Sum of the first terms exceed the term by
Sum of the and term is twice the sum of the first and second term
- $u_{3}+u_{4}=2\left(u_{1}+u_{2}\right)$ Remember to put the brackets around $u_{1}+u_{2}\,$
Sum of to term is times the sum of first terms
- $S_{14}-S_{6}=3S_{6}\,$
times the sum of the first terms is equal the sum of the next terms

Example

In an AP, the $3^{rd}\,$ term is twice the first term, and the sum of the first $4\,$ terms is $28\,$ . Find the first term and common difference.

- Analyze :
- $u_{3}=2u_{1}\qquad\qquad S_{4}=28$
- $a+2d=2a\qquad \frac{4}{2}\left(2a+3d\right)=28$
- $a=2d\frac{\quad}{}(1)\quad\;\; 2a+3d=14\frac{\quad}{}(2)$
- $\begin{align} &(1)\to(2): \\ & 4d+3d=14 \\ & 7d=14\\ & d=2, \therefore a=4\\ \end{align}$
- $\therefore \mbox{first term} = 4, \mbox{common difference} = 2$
- Check : $\underbrace{\overbrace{4,6,8}^{\times 2},10}_{Sum =28}$

Given S_n, find u_n

If we are given the formula for $S_{n}\,$ , how do we get back the formula for $u_{n}\,$ ? Well, first we see what is their relationship.

$\overbrace{\underbrace{u_{1}+u_{2}+\ldots+u_{n-1}}_{{\color{Red}??}}+{\color{Blue}u_{n}}}^{{\color{Blue}S_{n}}}$ $\overbrace{\underbrace{u_{1}+u_{2}+\ldots+u_{n-1}}_{{\color{Red}S_{n-1}}}+{\color{Blue}u_{n}}}^{{\color{Blue}S_{n}}}$
$\therefore u_{n}=$ $S_{n}-S_{n-1}\,$

Note

Nowhere did we use properties of AP, thus this formula applies for
$S_{1}\,=$ $u_{1}\,=$ $a\,$ (both AP & GP)

Example

Given $S_{n}=2n^{2}+n\,$ . Find $u_{3}\,$ & $u_{n}\,$ . Prove that $S_{n}\,$ is an arithmetic series.
- $S_{4}-S_{3}\,$
  $=\left[2\left(4\right)^{2}+4\right]-\left[2\left(3\right)^{2}+3\right]=15$
- $S_{n}-S_{n-1}\,$
  $=\left(2n^{2}+n\right)-\left[2\left(n-1\right)^{2}+\left(n-1\right)\right]$ Note that $S_{{\color{Red}n-1}}=2{\color{Red}\left(n-1\right)}^{2}+{\color{Red}\left(n-1\right)}$
  $=2n^{2}+n-\left(2n^{2}-4n+2+n-1\right)$
  $=2n^{2}+n-2n^{2}+3n-1\,$
  $=4n-1\,$
- Analyze :
- $\begin{align} & u_{n+1}-u_{n}=4\left[\left(n+1\right)-1\right]-\left(4n-1\right)\\ & \qquad \qquad =4\mbox{(constant)}\\ & \therefore S_{n}\;\mbox{is an arithmetic series} \end{align}$
- Check :

Inequalities

The one thing to remember here is that $n\,$ must be a positive integer.

Example

An AP has first term $1\,$ and common difference $4\,$ . Find the least value of $n\,$ such that

$u_{n}>1000\,$
- $\therefore 1+\left(n-1\right)4>1000$
- $\therefore \mbox{least value of }n\mbox{ is }251$
- Check : $u_{251}=1001,u_{250}=997\,$

Sum of Integers

Example : Find the sum of integers from $1\,$ to $500\,$ inclusive

i) which are odd
ii) which can be divided by $3\,$
iii) odd numbers that can be divided by $3\,$
iv) which are divisible by $3\,$ or $5\,$
v) which cannot be divided by $3\,$
vi) which cannot be divided by $3\,$ or $5\,$

Solution

i) which are odd
- Analyze :
- $\mbox{Let }S=\underbrace{1+3+5+\ldots+499}_{{\color{Red}\mbox{AP with}\;a=1,\; d=2,\; u_{n}=499}}$
- $\begin{align} & u_{n}=499 \\ & 1+\left(n-1\right)2=499 \\ & \therefore n=250 \\ \end{align}$
- $=62500\,$

ii) which can be divided by $3\,$
- Analyze :
- $\mbox{Let }S=3+6+9+\ldots+498\,$
- $\mbox{Let }S=\underbrace{3+6+9+\ldots+498}_{{\color{Red}\mbox{AP with}\;a=3,\; d=3,\; n=166}}$
- $=\frac{166}{2}\left(3+498\right)=41583$

iii) odd numbers that can be divided by $3\,$
- Analyze :
- $\mbox{Let }S=\underbrace{3+9+12+\ldots}_{\mbox{AP with}\;a=3,\; d=6}$
- $\begin{align} & u_{n} \le 500 \\ & 3+\left(n-1\right)6 \leq 500 \\ & n \le 83.4833 \\ & n=83, \therefore u_{n}=495 \\ \end{align}$
- $S=\frac{83}{2}\left(3+495\right)=20667$

iv) which are divisible by $3\,$ or $5\,$
- Analyze :
- $\begin{align} \mbox{Let } S_{a}&=3+6+9+\ldots+498 \\ S_{b} & = 5+10+15+\ldots+500\\ S_{c} & = 15+30+45+\ldots+495\\ \end{align}$ $\begin{align} \mbox{Let } S_{a}&=\underbrace{3+6+9+\ldots+498}_{{\color{Red}\mbox{AP with}\;a=3,\; d=3,\; n=166}} \\ S_{b} & = \underbrace{5+10+15+\ldots+500}_{{\color{Red}\mbox{AP with}\;a=5,\; d=5,\; n=100}}\\ S_{c} & = \underbrace{15+30+45+\ldots+495}_{{\color{Red}\mbox{AP with}\;a=15,\; d=15,\; n=33}}\\ \end{align}$
- $\mbox{Sum} =S_{a}+S_{b}-S_{c}\,$
- $=\frac{166}{2}\left(3+498\right)+\frac{100}{2}\left(5+500\right)$ $-\frac{33}{2}\left(15+495\right)$
- $=41583+25250-8415=58418\,$

v) which cannot be divided by $3\,$
- Analyze :
- $\begin{align} \mbox{Let } S_{a}&=\underbrace{1+2+3+\ldots+500}_{\mbox{AP with}\;a=1,\; d=1,\; n=500} \\ S_{b}&=\underbrace{3+6+9+\ldots+498}_{\mbox{AP with}\;a=3,\; d=3,\; n=166} \\ \end{align}$
- $\mbox{Sum} =S_{a}-S_{b}\,$
- $=\frac{500}{2}\left(1+500\right)-\frac{166}{2}\left(3+498\right)=83667$

vi) which cannot be divided by $3\,$ or $5\,$
- Analyze :

Exercise 1

Notes

As usual, go through all the notes and examples and make sure you understand all of it BEFORE attempting this, as well as memorize all the needed formulas.
As always with questions with lots of calculation, count carefully and deliberately, but don't overwrite too many calculation steps.
Questions that can be checked directly quickly should be checked. Questions that can't, try to check solution to simultaneous equation, etc, or just go through the workings once again to spot careless mistakes.

1) Prove that the following forms an AP

i) $\lg a,\lg ar,\lg ar^{2}\,$ where $a,r\,$ are constants
- $\begin{align} & u_{3}-u_{2}=\lg ar^{2}-\lg ar=\lg\frac{ar^{2}}{ar}=\lg r \\ & u_{2}-u_{1}=\lg ar-\lg a=\lg\frac{ar}{a}=\lg r \\ & \therefore u_{3}-u_{2}= u_{2}-u_{1}\\ & \therefore \mbox{It is an AP} \end{align}$

ii) $u_{n}=1-3n\,$ . Find also first term and common difference.
$a=-2, d=3\,$
- $\begin{align} u_{n+1}-u_{n} & =\left[1-3\left(n+1\right)\right]-\left(1-3n\right) \\ & = 1-3n-3-1+3n \\ & = -3\mbox{(constant)}\\ \end{align}$
- $\begin{align} & \therefore \mbox{It is an AP} \\ & a=u_{1}=1-3=-2\\ & d=-3\\ \end{align}$
- Check : $u_{n}=-2+\left(n-1\right)\left(-3\right)=-2-3n+3=1-3n$

2) Find sum of the following series

i) $-7-2+3+\ldots$ until the $30^{th}\,$ term
$1965\,$
- $\underbrace{-7-2+3+\ldots}_{\mbox{AP with}\;a=-7,\; d=5,\; n=30}$
- $\begin{align} S_{30} & = \frac{30}{2}\left[-14+29\left(5\right)\right]\\ & = 1965 \end{align}$

ii) $3+7+11+\ldots+491$
$30381\,$
- $\underbrace{3+7+11+\ldots+491}_{\mbox{AP with}\;a=3,\; d=4,\; u_{n}=491}$
- $\begin{align} & u_{n}=491 \\ & \therefore 3+\left(n-1\right)4=491\\ & n=123 \\ & S_{123} = \frac{123}{2}\left(3+491\right)\\ & \qquad = 30381 \end{align}$

3) Find the first term and common difference in the following AP if

i) The $9^{th}\,$ term is $5\,$ times the $2^{nd}\,$ term and sum of the first $6\,$ terms is $78\,$ . Find also the sum of $7^{th}\,$ to $12^{th}\,$ term.
$a=3, d=4 ; 222\,$
- $\begin{array}{ll} u_{9}=5u_{2} & S_{6}=78 \\ a+8d=5\left(a+d\right) & \frac{6}{2}\left(2a+5d\right)=78\\ 3d=4a & 2a+5d=26 \\ 4a=3d\frac{\quad}{}(1) & 4a+10d=52\frac{\quad}{}(1) \end{array}$
- $\begin{align} & (1)\to(2): \\ & 3d+10d=52\\ & 13d=52 \\ & d=4, 4a=12, \therefore a=3\\ & \therefore a=3, d=4 \end{align}$
- $\begin{align} \mbox{Sum} & =S_{12}-S_{6} \\ & = \frac{12}{2}\left[6+11\left(4\right)\right]-\frac{6}{2}\left[6+5\left(4\right)\right]\\ & = 222 \end{align}$
- Check : $u_{9}=35, u_{2}=7, S_{6}=78\,$

ii) The $5^{th}\,$ term is $-18\,$ and the sum of the $3^{rd}\,$ and $4^{th}\,$ term is thrice the sum of the first two terms. Find also the sum of the $7^{th}\,$ and $8^{th}\,$ term.
$a=-2, d=-4 ; -56\,$
- $\begin{array}{ll} u_{5}=-18 & u_{3}+u_{4}=3\left(u_{1}+u_{2}\right) \\ a+4d=-18\frac{\quad}{}(1) & \left(a+2d\right)+\left(a+3d\right)=3\left(a+a+d\right)\\ & 2a+5d=6a+3d \\ & 2d=4a \\ & d=2a\frac{\quad}{}(2) \\ \end{array}$
- $\begin{align} & (2)\to(1): \\ & a+8a=-18\\ & 9a=-18, \therefore a=-2,d=-4 \\ & u_{7}+u_{8}=\left[-2+6\left(-4\right)\right]+\left[-2+7\left(-4\right)\right]\\ & \qquad \quad =-56\end{align}$
- Check : $\overbrace{-2,-6}^{-8},\overbrace{-10,-14}^{-24},-18\,$

iii) First term is $15\,$ and the sum of the first $6\,$ terms is equal to the sum of the next $4\,$ terms.
$d=2\,$
- $\begin{array}{ll} a=15 \quad & S_{6}=S_{10}-S_{6} \\ & 2S_{6}=S_{10} \\ & 2\left[\frac{6}{2}\left(30+5d\right)\right]=\frac{10}{2}\left(30+9d\right)\\ & 180+30d=150+45d \\ & 15d=30 \\ & \therefore d=2\\ \end{array}$
- Check : $\overbrace{15,17,19,21,23,25}^{120},\overbrace{27,29,31,33}^{120}$

4) In an AP $\left(a=5,d=-1\right)$ , twice the sum of first $3n\,$ terms equals the sum of the next $n\,$ terms. Find value of $n\,$ .

$n=5\,$
- $\begin{align} & 2S_{3n}=S_{4n}-S_{3n}\\ & 3S_{3n}=S_{4n} \\ & 3\left\{\frac{3n}{\cancel{2}}\left[10+\left(3n-1\right)\left(-1\right)\right]\right\}=\frac{4n}{\cancel{2}}\left[10+\left(4n-1\right)\left(-1\right)\right]\\ & 9n\left(11-3n\right)=4n\left(11-4n\right) \\ & n\left(99-27n\right)-n\left(44-16n\right)=0 \\ & n\left(55-11n\right)=0 \\ & n=0\mbox{(rejected) }\mbox{ or } 11n =55\\ &\therefore n=5 \\ \end{align}$

5) Prove that $S_{n}=4n-n^{2}\,$ forms an arithmetic series and find its find first term and common difference.

$a=3, d=-2\,$
- $\begin{align} u_{n} & = S_{n}-S_{n-1}\\ & = \left(4n-n^{2}\right)-\left[4\left(n-1\right)-\left(n-1\right)^{2}\right]\\ & = 4n-n^{2}-\left(4n-4-n^{2}+2n-1\right)\\ & = 5-2n \\ \end{align}$
- $\begin{align} u_{n+1}-u_{n}& = \left[5-2\left(n+1\right)\right]-\left(5-2n\right)\\ & = -2\mbox{(constant)} \end{align}$
- $\begin{align} &\therefore S_{n} \mbox{ is an arithmetic series}\\ & a=S_{1}=4-1=3, d=-2 \end{align}$
- Check : $S_{n}=\frac{n}{2}\left[6+\left(n-1\right)\left(-2\right)\right]=\frac{n}{2}\left(8-2n\right)=4n-n^{2}$

6)a) In an AP $\left(a=2,d=3\right)$ , find least value of $n\,$ such that

i) $u_{n}>1000\,$
$n=334\,$
- $\begin{align} u_{n}& >10000\\ 2+\left(n-1\right)3 & >1000\\ n &>333\frac{2}{3}\\ \end{align}$
- $\therefore \mbox{least value of }n\mbox{ is }334$
- Check : $u_{334}=1001\,$

b) In an AP $\left(a=500,d=-7\right)$ , find least such that

i) $u_{n}<200\,$
$n=44\,$
- $\begin{align} u_{n}& <200\\ 500+\left(n-1\right)\left(-7\right) & <200\\ n &>43\frac{6}{7}\\ \end{align}$
- $\therefore \mbox{least value of }n\mbox{ is }44$
- Check : $u_{44}=199\,$

7) Find sum of integers $1\,$ to $750\,$ inclusive

i) which are odd
$140625\,$
- $\begin{align} & \mbox{Let } S =\underbrace{1+3+5+\ldots+749}_{\mbox{AP with}\;a=1,\; d=2,\; u_{n}=749}\\ & u_{n}=749 \\ & 1+\left(n-1\right)2=749 \\ & n=375 \\ & \therefore S=\frac{375}{2}\left(1+749\right)\\ & \qquad =140625 \end{align}$

ii) which are divisible by $3\,$
$94125\,$
- $\begin{align} & \mbox{Let } S =\underbrace{3+6+9+\ldots+750}_{\mbox{AP with}\;a=3,\; d=3,\; n=250}\\ & \therefore S=\frac{250}{2}\left(3+750\right)\\ & \qquad =94125 \end{align}$

iii) which are even numbers that can be divided by $7\,$
$20034\,$
- $\begin{align} & \mbox{Let } S =\underbrace{14+28+42+\ldots+742}_{\mbox{AP with}\;a=14,\; d=14,\; n=53}\\ & \therefore S=\frac{53}{2}\left(14+742\right)\\ & \qquad =20034 \end{align}$

iv) which are divisible by $3\,$ or $4\,$
$141001\,$
- $\begin{align} \mbox{Let } S_{a}&=\underbrace{3+6+9+\ldots+750}_{\mbox{AP with}\;a=3,\; d=3,\; n=250} \\ S_{b} & = \underbrace{4+8+12+\ldots+748}_{\mbox{AP with}\;a=4,\; d=4,\; n=187}\\ S_{c} & = \underbrace{12+24+36+\ldots+744}_{\mbox{AP with}\;a=11,\; d=11,\; n=62}\\ & \mbox{Sum} = S_{a}+S_{b}-S_{c}\\ & \qquad =\frac{250}{2}\left(3+750\right)+\frac{187}{2}\left(4+748\right)-\frac{62}{2}\left(12+744\right)\\ & \qquad =94125+70312-23436=141001 \end{align}$

v) which cannot be divided by $5\,$
$225000\,$
- $\begin{align} \mbox{Let } S_{a}&=\underbrace{1+2+3+\ldots+750}_{\mbox{AP with}\;a=1,\; d=1,\; n=750} \\ S_{b} & = \underbrace{5+10+15+\ldots+750}_{\mbox{AP with}\;a=5,\; d=5,\; n=150}\\ & \mbox{Sum} = S_{a}-S_{b}\\ & \qquad =\frac{750}{2}\left(1+750\right)-\frac{150}{2}\left(5+750\right)\\ & \qquad =281625-56625=225000 \end{align}$

vi) which cannot be divided by $5\,$ or $7\,$
$192639\,$
- $\begin{align} \mbox{Let } S_{a}&=\underbrace{5+10+15+\ldots+750}_{\mbox{AP with}\;a=5,\; d=5,\; n=150} \\ S_{b} & = \underbrace{7+14+21+\ldots+749}_{\mbox{AP with}\;a=7,\; d=7,\; n=107}\\ S_{c} & = \underbrace{35+70+105+\ldots+735}_{\mbox{AP with}\;a=35,\; d=35,\; n=21}\\ & \mbox{Sum of numbers that can be divided by 5 or 7}=S_{a}+S_{b}-S_{c}\\ & \qquad =\frac{150}{2}\left(5+750\right)+\frac{107}{2}\left(7+749\right)-\frac{21}{2}\left(35+735\right)\\ & \qquad =566255+40446-8085=88986 \\ S_{d} & = \underbrace{1+2+3+\ldots+750}_{\mbox{AP with}\;a=1,\; d=1,\; n=750}\\ & \mbox{Sum of numbers that cannot be divided by 5 or 7}=S_{d}-88986 \\ & \qquad = \frac{750}{2}\left(1+750\right)-88986 \\ & \qquad = 192639\\ \end{align}$

Arithmetic Geometric Progression Part1

Contents

Notes

Learning Objectives (Syllabus)

Prior Knowledge

Arithmetic Progression

Notation

u_n for Arithmetic Progression

Examples

Definition of AP

Examples

Definition(Formula)

Arithmetic Mean

S_n for Arithmetic Progression

Prove

Formula

Examples

Sum NOT from the first term

Interpreting Information

Given S_n, find u_n

Inequalities

Sum of Integers

Exercise 1

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Arithmetic Geometric Progression Part1

Contents

Notes

Learning Objectives (Syllabus)

Prior Knowledge

Arithmetic Progression

Notation

un for Arithmetic Progression

Examples

Definition of AP

Examples

Definition(Formula)

Arithmetic Mean

Sn for Arithmetic Progression

Prove

Formula

Examples

Sum NOT from the first term

Interpreting Information

Given Sn, find un

Inequalities

Sum of Integers

Exercise 1

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u_n for Arithmetic Progression

S_n for Arithmetic Progression

Given S_n, find u_n