Arithmetic Geometric Progression Part1

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Contents


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

  • 2,5,8,11,\, 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}
  • 10,3,-4,\, -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,x+y,x+2y,\, 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 add it one by one, we end up with a series.


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

  • S_{2n}\, Hmmm... should we go u_{2}+u_{4}+u_{6}+\ldots\,? No. 2n\, just tell us which term to add until, the sum ALWAYS starts from first term.
    • S_{2n}=\, u_{1}+u_{2}+u_{3}+\ldots+ u_{2n}\,

un 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=\, common difference It might seem a bit weird to call this difference when we actually add it from term to term, a \xrightarrow{+d} (a+d) \xrightarrow{+d} (a+2d) \xrightarrow{+d} (a+3d) but it is correct as we get the value d\, back when we subtract one term from the next.

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,

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

Examples

  • u_{n+1}=\, a+nd\,
    • Its because {\color{Red}\left(n+1\right)}-1=n
  • u_{n-1}=\, a+\left(n-2\right)d\,
  • u_{2n}=\, 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 difference between successive terms is constant.

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. To say common difference implies that we already assume it is the same(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 :
      • Keep in mind we need to prove that difference between successive terms are constant
      • Since there are only 3 terms, there are only two differences, and we just need to prove that there are the same
    • u_{3}-u_{2}=11-7=4\,
    • u_{2}-u_{1}=7-3=4\,
    • \therefore \mbox{it is an AP} We can also go and write "Since the difference between successive terms are constant, it is an AP", but I think it should be OK not to write in this case.
    • Note :
      • It is WRONG to write the prove directly as
        • u_{3}-u_{2}=u_{2}-u_{1}\, OR
        • d=11-7=7-3\, as both means that we are already assuming it to be an AP. We need to prove it is an AP. Both statement, though are correct statements, will result in zero marks.

How about the second one?

    • Analyze:
      • Hmmm... we don't have the terms, only a formula for it. So what do we do?
      • Perhaps like this
        • \begin{align} & u_{n}=4n-1 \\ & \therefore u_{1}=4-1=3 \\ & u_{2}=8-1=7 \\ & u_{3}=12-1=11 \\ & u_{3}-u_{2}=11-7=4 \\ & u_{2}-u_{1}=7-3=4 \\ & \mbox{Thus, it is an AP.} \end{align}
        • Correct? Yes, but ZERO marks. But why? The numbers are exactly like the first question. Because the question asks us to prove that u_{n}=4n-1\, is an AP, meaning we have to prove ALL of the differences between successive terms are equal, not just the first three terms. At best, we can consider the above answer to be an incomplete one, but since we won't be able to prove for all using this way (even if you can write out till the million-th term, it will still be incomplete), no marks will be given.
      • How about this
        • 4n-1=a+\left(n-1\right)d
        • And then we proceed to solve for a,d\,, and thus say it is an AP? Zero marks. Take note that the questions asks us to prove it is an AP, and then only find the first term and common difference. By using the formula for AP, we have already assumed that it is an AP. Once again, remember this -> Never assume something that we are supposed to prove.
      • So we know we would still need to prove the differences are common, and we need to prove
        • \begin{align} & u_{2}-u_{1}= \\ & u_{3}-u_{2}= \\ & u_{4}-u_{3}= \\ & u_{5}-u_{4}= \\ & \ldots \\ & u_{100}-u_{99}= \\ & \ldots \\ & u_{100000}-u_{99999}= \\ & \ldots \\ \end{align}
        • are all the same. But since all of them have the same pattern, we just need to prove it once using a general formula to represent the difference between successive terms. Since we are taking u_{{\color{Red}3}}-u_{{\color{Blue}2}}\,, u_{{\color{Red}4}}-u_{{\color{Blue}3}},\ldots, u_{{\color{Red}100}}-u_{{\color{Blue}99}}, thus it will be u_{{\color{Red}??}}-u_{{\color{Blue}n}} u_{{\color{Red}n+1}}-u_{{\color{Blue}n}}
        • Thus, we now know that we would need to prove u_{n+1}-u_{n}\, is constant. Note that we can also use u_{n}-u_{n-1}\,, but plus is always easier than minus, and u_{n+1}-u_{n}\, is more consistent with formula for u_{n}\, as both starts with n=1\,
    • u_{n+1}-u_{n}=\, \left[4\left(n+1\right)-1\right]-4n-1
    • =4n+3-4n+1\,
    • =4\;\mbox{(constant)}\, Note that we must write the "(constant)" here.
    • \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
      • You might also ask, what will be considered not constant? if you get something like n+1, 2n\, and so on
    • \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 c-b=b-a\; This is because we know the difference is the same
  • \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.

Sn for Arithmetic Progression

Example

  • 1+2+3+\ldots +100 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 u_{n}=\, a+\left(n-1\right)d
      • To make things easier to write, however, we will use l\, to represent the last term (u_{n}\,)
    • Note that since we need to "add to" the first 3 terms, we would also need to write out the last three terms
      • The term before the last term is l-d\,
        • {\color{Red}??}\xrightarrow{{\color{Blue}+d}}l \therefore {\color{Red}??}=l-{\color{Blue}d}
      • And the one before that l-2d\,
    • 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 add both of them together, pairing the terms
      • Not that the left hand side is 2S_{n}\,
      • Each pair will result in \left(a+l\right). Note that if you don't get the same result for every pair, then something is wrong.
      • (1)+(2):\,
      • 2S_{n}=\left(a+l\right)+\left(a+l\right)+\left(a+l\right)+\ldots+\left(a+l\right)+\left(a+l\right)+\left(a+l\right)
      • How many terms?
        • 2S_{n}=\underbrace{\left(a+l\right)+\left(a+l\right)+\left(a+l\right)+\ldots+\left(a+l\right)+\left(a+l\right)+\left(a+l\right)}_{n\;terms}
    • Thus 2S_{n}=\, n\left(a+l\right)
    • \therefore S_{n}=\frac{n}{2}\left(a+l\right)
  • Thus we have S_{n}=\frac{n}{2}\left(a+l\right)
    • 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(a+l\right) Use this when we have last term. It's easier to count.(We will still need n\,, however)
  • 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 : We don't have the last term, so we can't use the simpler formula.
      • Also note that the question never states that it is an arithmetic series. We should state that we understand it is an arithmetic series (and might as well list down the known quantities of it)
    • \underbrace{7+4+1+\ldots u_{20}}_{\mbox{AP with}\;a=7,\; d=-3,\; n=20}
    • S_{20}=\, \frac{20}{2}\left[{\color{Red}14}+{\color{Blue}19}\left(-3\right)\right] Don't make careless mistakes when putting in the values of {\color{Red}2a} and {\color{Blue}n-1}
    • =-430\,


  • 2+5+8+\ldots+296
    • Analyze : We have the last term, so we can use the simpler formula. But we still would need to find n\, (how many terms there are) first, and we will find it from value of u_{n}\, (the last term)
    • \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

  • 5^{th}\, term to 8^{th}\, 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 subtract the terms in front which we do not want
        • \overbrace{\underbrace{{\color{Red}u_{1}+u_{2}+u_{3}+u_{4}}}_{{\color{Red}??}}+{\color{Blue}u_{5}+u_{6}+u_{7}+u_{8}}}^{S_{8}}
        • \overbrace{\underbrace{{\color{Red}u_{1}+u_{2}+u_{3}+u_{4}}}_{{\color{Red}S_{4}}}+{\color{Blue}u_{5}+u_{6}+u_{7}+u_{8}}}^{S_{8}}
      • 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.

  • 5^{th}\, term is 6\, and sum of first 5\, terms is 50\,
    • u_{5}=6,\;S_{5}=50\,
  • The difference between the 7^{th}\, and 3^{rd}\, term is 2\, times the 5^{th}\, term
    • u_{7}-u_{3}=2u_{5}\, (assuming here d\, is positive, so that u_{7}>u_{3}\,)
  • Sum of the first 3\, terms exceed the 5^{th}\, term by 10\,
    • S_{3}-u_{5}=10\, When we say a\, exceeds(larger than) b\, by 2\,, that means their difference is 2\,
  • Sum of the 3^{rd}\, and 4^{th}\, 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 7^{th}\, to 14^{th}\, term is 3\, times the sum of first 6\, terms
    • S_{14}-S_{6}=3S_{6}\,
  • 3\, times the sum of the first 2n\, terms is equal the sum of the next n\, terms
    • 3S_{2n}=S_{3n}-S_{2n}\, Take your time when seeing these type of questions. Remember, an extra 20 seconds or so to make sure this is correct can prevent big loss in time and marks


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 : Certainly not a difficult question, but that means more the reason to be careful as to not lose any marks unnecessarily. As the values are small, it is also possible to check directly that our answers satisfy the conditions given
    • 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 Sn, find un

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.

  • {\color{Blue}S_{n}}\, =u_{1}+u_{2}+\ldots+{\color{Blue}u_{n}}
    • If we want {\color{Blue}u_{n}} only, that means we don't want the terms in front. The one just before {\color{Blue}u_{n}} is u_{n-1}\,
  • \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 ALL types of sequences (AP, GP, or any other type)
  • 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.
    • u_{3}\,= S_{4}-S_{3}\,
      • =\left[2\left(4\right)^{2}+4\right]-\left[2\left(3\right)^{2}+3\right]=15
    • u_{n}\,= 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 : How do we go to prove that it is an AP?
      • Can we go find its a, d\, like below?
        • \begin{align} & u_{1} = S_{1} \\ & \qquad = 2+1 =3 \\ & u_{2} = S_{2}-S_{1} \\ & \qquad = 10-3 =7 \\ & d= u_{2}-u_{1}=7-3=4\\ & \therefore \mbox{it is an arithmetic series}\\ \end{align}
        • NO. This would only be acceptable if the question told us that it is an arithmetic series, and asks us to find first term and common difference. But we are asked to prove that it is. Thus, we still have to use back what we learn earlier
    • \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 : We can check that we get back the given formula. a=u_{1}=3, S_{n}=\frac{n}{2}\left[6+\left(n-1\right)4\right] =\frac{n}{2}\left(4n+2\right)=2n^{2}+n

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\,
  • u_{n}>1000\,
    • \therefore 1+\left(n-1\right)4>1000
    • n>250\frac{3}{4}\, Note that though normally we will write \frac{1003}{4}, we are looking for an integer here
    • \therefore \mbox{least value of }n\mbox{ is }251
    • Check : u_{251}=1001,u_{250}=997\,

  • S_{n}>1000\,
  • S_{n}>1000\,
    • \frac{n}{2}\left[2+\left(n-1\right)4\right]>1000\,
    • n+2n^{2}-2n>1000\,
    • 2n^{2}-n-1000>0\, Using the calculator, we will find that the roots are real irrational roots (surd). In normal situation, we would then need to use the quadratic formula to get the exact value of the roots. But since we are only looking for an integer, it should be ok just to use the decimals from the calculator
    • \left(n-22.61\right)\left(n+22.11\right)>0
    • MT-APGP-Ex-1.pngMedia:MT-APGP-Ex-1.dia
    • n<-22.11 \mbox{ or } n>22.61\,
    • \therefore \mbox{least value of }n\mbox{ is }23
    • Check : S_{23}=1035,S_{22}=946\,

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

  • First note that "inclusive" means including 1\, and 500\, itself. Mathematically speaking, when we say "between a\, and b\,", it does NOT include a\, and b\,
  • i) which are odd
    • Analyze : Obviously, since we know we are doing AP now, we know this is an arithmetic series. But generally speaking, if we can't see what type of sequence it is, we should list down the terms first
    • \mbox{Let }S=1+3+5+\ldots +499\, We can now see it is an AP. We have the last term.
    • \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}
    • \therefore S=\frac{250}{2}\left(1+499\right) Be careful not to substitute wrongly for S_{n}=\frac{{\color{Blue}n}}{2}\left(a+{\color{Red}l}\right)
    • =62500\,

  • ii) which can be divided by 3\,
    • Analyze : The number wills be 3+6+9+\ldots. What will be the last term? We need a number smaller than 500\, that can be divided by 3\,
      • A fast method to do that is just to take your calculator and count 500\div 3={\color{Blue}166}.66666, and take that {\color{Blue}166} and multiply back 3\,, that is 166 \times 3=498
      • Also, since we can see clearly that the terms have the form of 3\times {\color{Blue}1} , 3\times {\color{Blue}2}, 3\times {\color{Blue}3},\ldots, 3\times {\color{Blue}166} , we can see directly that n\, =166\, and we wouldn't need to do the usual working to find it out
    • \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 : We only want the multiples of 3 which are odd, 3+\cancel{6}+9+\cancel{12}+15+\ldots. We can see that it will still be an AP
      • The last term takes a bit of calculation to get, you can directly use calculator for try and error or you can work it out using u_{n}\le 500\,
    • \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}
    • We can of course skip finding u_{n}\, and just use the other S_{n}\, formula, but I did it here to make sure we got the correct n\,
    • S=\frac{83}{2}\left(3+495\right)=20667

  • iv) which are divisible by 3\, or 5\,
    • Analyze :
      • Numbers which are divisible by 3\, or 5\,: 3,5,6,9,10,12,\ldots But we can see that this is not an AP, or have any pattern that will enable us to calculate it by formula
      • Numbers which are divisible by 3\,: 3,6,9,12,\ldots
      • Numbers which are divisible by 5\,: 5,10,15,\ldots
      • Taken separately, these two are APs, so it seems that we will take the total of both sums. However, on closer look
        • Numbers which are divisible by 3\, : 3,6,9,12,{\color{Red}15},\ldots
        • Numbers which are divisible by 5\, : 5,10,{\color{Red}15},\ldots
        • We see that there are numbers that are repeated, and thus will be counted twice if we just add the two sums together. But we should only count it once, thus we need it subtract it one time after we add the two sums
        • What other numbers? The number must appear on both sums, meaning can be divided by both 3\, AND 5\,, thus can be divided by 15\,. 15,30,45,\ldots
      • Since there are 3 sums, we label them with S_{a},S_{b},S_{c}\, (avoid using numbers as it can be confusing)
    • \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 : 1+2+\cancel{3}+4+5+\cancel{6}+7+8+\cancel{9}+\ldots, we want all numbers except those that that can be divided by 3\, , thus we take the sum of all numbers minus sum of those numbers that can be divided by 3\,
    • \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 : As always, when faced with complicated questions, let's see what basic parts of the question which we already know how to do first
      • which cannot be divided by 3\, or 5\,
        • divided by 3\, or 5\, : We know how to do this already. In fact, its exactly the same as iv)
        • cannot be divided : We know how to deal with "cannot" in question v)
        • Thus, we see that we will figure out the sum in two steps. We can, however choose to leave the actual calculations till the end.
        • Note : Trying to deal with it in one step will be unwise, and doesn't necessarily show extra intelligence. Wise men don't complicate the simple. They simplify the complicated . =)
    • So lets deal with the can divided by 3\, or 5\, first (in this particular question, we can actually just take the value from iv), but let's assume iv) is not asked so that we can practice writing the whole solution)
      • \begin{align} \mbox{Let } S_{a}&=\underbrace{3+6+9+\ldots+498}_{\mbox{AP with}\;a=3,\; d=3,\; n=166} \\ S_{b} & = \underbrace{5+10+15+\ldots+500}_{\mbox{AP with}\;a=5,\; d=5,\; n=100}\\ S_{c} & = \underbrace{15+30+45+\ldots+495}_{\mbox{AP with}\;a=15,\; d=15,\; n=33}\\ \end{align}
      • \mbox{Sum of numbers that can be divided by 3 or 5}\, =S_{a}+S_{b}-S_{c}\,
      • \begin{align} &=\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\end{align}
    • And then only deal with the cannot
      • \mbox{Let } S_{d}=\underbrace{1+2+3+\ldots+500}_{\mbox{AP with}\;a=1,\; d=1,\; n=500}
      • \mbox{Sum of numbers that cannot be divided by 3 or 5}\, =S_{d}-58418\,
      • =\frac{500}{2}\left(1+500\right)-58418=66832

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}
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