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Symbols and Formulas: Arithmetic Sequence
$AS_n$ = $nth$ term of an Arithmetic Sequence
$a$ = first term
$p$ = last term
$d$ = common difference
$n$ = number of terms
$SAS_n$ = sum of the first $n$ terms of an Arithmetic Sequence
$
(1.)\:\: AS_n = a + d(n - 1) \\[5ex]
(2.)\:\: AS_n = vn + w \:\:where\:\: v = d \:\:and\:\: w = a - d \\[5ex]
(3.)\:\: p = a + d(n - 1) \\[5ex]
(4.)\:\: SAS_n = \dfrac{n}{2}(a + AS_n) \\[7ex]
(5.)\:\: SAS_n = \dfrac{n}{2}(a + p) \\[7ex]
(6.)\:\: SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[7ex]
(7.)\:\: n = \dfrac{2 * SAS_n}{a + p} \\[7ex]
(8.)\:\: n = \dfrac{p - a + d}{d} \\[7ex]
(9.)\:\: n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 + 8d*SAS_n}}{2d} \\[7ex]
(10.)\;\; d = \dfrac{(p - a)(p + a)}{2 * SAS_n - p - a}
$
Symbols and Formulas: Geometric Sequence
$GS_n$ = $nth$ term of a Geometric Sequence
$a$ = first term
$p$ = last term
$r$ = common ratio
$n$ = number of terms
$SGS_n$ = sum of the first $n$ terms of a Geometric Sequence
$S_{\infty}$ = sum to infinity of a Geometric Sequence
(3.) The third and sixth terms of a Geometric Progression (G.P) are $\dfrac{1}{4}$ and $\dfrac{1}{32}$
respectively.
Find the:
(i) first term and the common ratio
(ii) seventh term
(4.) The fifth term of an Arithmetic Progression (A.P) is 11 and the eighth term is 20
Find the:
(i) 12th term
(ii.) Sum of the first 12 terms
$
AS_n = a + d(n - 1) \\[3ex]
AS_{5} = a + 4d \\[3ex]
11 = a + 4d \\[3ex]
a + 4d = 11...eqn.(1) \\[3ex]
AS_{8} = a + 7d \\[3ex]
20 = a + 7d \\[3ex]
a + 7d = 20...eqn.(2) \\[3ex]
eqn.(2) - eqn.(1) \implies \\[3ex]
(a + 7d) - (a + 4d) = 20 - 11 \\[3ex]
a + 7d - a - 4d = 9 \\[3ex]
3d = 9 \\[3ex]
d = \dfrac{9}{3} \\[5ex]
d = 3 \\[3ex]
From\;\; eqn.(1) \\[3ex]
a = 11 - 4d \\[3ex]
a = 11 - 4(3) \\[3ex]
a = 11 - 12 \\[3ex]
a = -1 \\[3ex]
(i) \\[3ex]
AS_{12} = a + 11d \\[3ex]
= -1 + 11(3) \\[3ex]
= -1 + 33 \\[3ex]
= 32 \\[3ex]
(ii) \\[3ex]
SAS_n = \dfrac{n}{2}(a + AS_n) \\[5ex]
SAS_{12} = \dfrac{12}{2}(-1 + AS_{12}) \\[5ex]
= \dfrac{12}{2}(-1 + 32) \\[5ex]
= 6(31) \\[3ex]
= 186
$
(5.) The second, fourth and sixth terms of an Arithmetic Progression (AP) are $x - 1$, $x + 1$, and 7
respectively.
Find the
(i) common difference;
(ii) first term;
(iii) value of x
$
(i) \\[3ex]
AS_n = a + d(n - 1) \\[3ex]
AS_2 = a + d = x - 1...eqn.(1) \\[3ex]
AS_4 = a + 3d = x + 1...eqn.(2) \\[3ex]
AS_6 = a + 5d = 7...eqn.(3) \\[3ex]
eqn.(2) - eqn.(1) \implies (a + 3d) - (a + d) = (x + 1) - (x - 1) \\[3ex]
a + 3d - a - d = x + 1 - x + 1 \\[3ex]
2d = 2 \\[3ex]
d = \dfrac{2}{2} \\[5ex]
d = 1 \\[3ex]
(ii) \\[3ex]
From\:\: eqn.(3) \\[3ex]
a + 5d = 7 \\[3ex]
a = 7 - 5d \\[3ex]
Subst.\:\: d = 1 \\[3ex]
a = 7 - 5(1) \\[3ex]
a = 7 - 5 \\[3ex]
a = 2 \\[3ex]
(iii) \\[3ex]
From\:\: eqn.(1) \\[3ex]
a + d = x - 1 \\[3ex]
x - 1 = a + d \\[3ex]
x = a + d + 1 \\[3ex]
Subst.\:\: a = 2 \:\:and\:\: d = 1 \\[3ex]
x = 2 + 1 + 1 \\[3ex]
x = 4
$
(6.) The 3rd, 4th and 5th terms of an exponential sequence are x, y, and z respectively.
Which of the following statements is true?
(7.) The difference between the third and first terms of a Geometric Progression (G.P) is 42.
If the fourth term is greater than the second term by 168, find the:
(i) first term
(ii) fourth term
of the progression.
(9.) The sum of the first ten terms of an Arithmetic Progression (A.P) is 130
If the fifth term is 3 times the first term, find the:
(a.) common difference
(b.) first term
(c.) number of terms of the A.P if the last term is 28
$
(a.) \\[3ex]
SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex]
SAS_{10} = \dfrac{10}{2}[2a + d(10 - 1)] \\[5ex]
130 = 5[2a + 9d] \\[3ex]
\dfrac{130}{5} = 2a + 9d \\[3ex]
26 = 2a + 9d \\[3ex]
2a + 9d = 26...eqn.(1) \\[3ex]
AS_n = a + d(n - 1) \\[3ex]
AS_5 = a + d(5 - 1) \\[3ex]
AS_5 = a + 4d \\[3ex]
But\;\; AS_5 = 3 * a \\[3ex]
\implies 3a = a + 4d \\[3ex]
3a - a = 4d \\[3ex]
2a = 4d \\[3ex]
\dfrac{2a}{2} = \dfrac{4d}{2} \\[5ex]
a = 2d...eqn.(2) \\[3ex]
Subst.\;\;eqn.(2)\;\;into\;\;eqn.(1) \\[3ex]
2(2d) + 9d = 26 \\[3ex]
4d + 9d = 26 \\[3ex]
13d = 26 \\[3ex]
d = \dfrac{26}{13} \\[5ex]
d = 2 \\[5ex]
(b.) \\[3ex]
Subst.\;\; d = 2\;\;into\;\; eqn.(2) \\[3ex]
a = 2d \\[3ex]
a = 2(2) \\[3ex]
a = 4 \\[5ex]
(c.) \\[3ex]
last\;\;term = p = 28 \\[3ex]
n = \dfrac{p - a + d}{d} \\[5ex]
n = \dfrac{28 - 4 + 2}{2} \\[5ex]
n = \dfrac{26}{2} \\[5ex]
n = 13
$
(10.) If $(3 - x),\;\; 6,\;\; (7 - 5x)$ are consecutive terms of a Geometric Progression with constant ratio
r > 0, find the:
(i) values of x;
(ii) common ratio
(11.) Find the sum of the Arithmetic Progression (AP) $1, 3, 5, ..., 101$
This is a sequence of odd numbers
So, we are asked to find the sum of the first $n$ odd numbers
We need to find the number of terms first
We can find $n$ in two ways.
Mentally
There are 5 oddnumbers from 1 – 10 (1, 3, 5, 7, 9)
There are 6 oddnumbers from 1 – 11 (1, 3, 5, 7, 9, 11)
Therefore, there are 50 odd numbers from 1 – 100
So, there are 51 odd numbers from 1 – 101
$
n = 51 \\[3ex]
\underline{By\:\: Formula} \\[3ex]
a = 1 \\[3ex]
p = 101 \\[3ex]
d = 3 - 1 = 2 \\[3ex]
n = \dfrac{p - a + d}{d} \\[5ex]
n = \dfrac{101 - 1 + 2}{2} \\[5ex]
n = \dfrac{102}{2} \\[5ex]
n = 51 \\[3ex]
$
We can also find $SAS_n$ in two ways.
You may use any of the formulas for $SAS_n$
(12.) The first term of an Arithmetic Progression (A.P) is 31 and the common difference is 9
Show that the nth term is 9n + 22
Hence, find the 20th term.
(13.) The nth term of a sequence is $5 + \dfrac{2}{3^{n - 2}}$ for $n \ge 1$.
What is the sum of the fourth and fifth terms?
Leave your answers in the form $\dfrac{x}{y}$ where x and y are integers.
(15.) If $\dfrac{1}{2}$, $\dfrac{1}{x}$, $\dfrac{1}{3}$ are successive terms of an arithmetic progression
(A.P),
show that $\dfrac{2 - x}{x - 3} = \dfrac{2}{3}$
(19.) The heights of students in a class form an Arithmetic Progression (A.P.) such that the tallest student
is 1.90 m.
The common difference in their heights is 0.05 m and the sum of their heights is 23.25 m.
(a.) How many students are in the class?
(b.) Find the height of the shortest student.
Let:
height of the tallest student = last term = p
height of the shortest student = first term = a
commom difference = d
number of students in the class = number of terms = n
sum of the arithmetic sequence = SASn
$
p = 1.90\;m \\[3ex]
d = 0.05\;m \\[3ex]
SAS_n = 23.25 \\[5ex]
p = a + d(n - 1) \\[3ex]
1.9 = a + 0.05(n - 1) \\[3ex]
a + 0.05n - 0.05 = 1.9 \\[3ex]
a + 0.05n = 1.9 + 0.05 \\[3ex]
a + 0.05n = 1.95 ...eqn.(1) \\[5ex]
SAS_n = \dfrac{n(a + p)}{2} \\[5ex]
23.25 = \dfrac{n(a + 1.9)}{2} \\[5ex]
n(a + 1.9) = 2(23.25) \\[3ex]
an + 1.9n = 46.5 ...eqn.(2) \\[5ex]
\underline{\text{Substitution Method}} \\[3ex]
From\;\;eqn.(1) \\[3ex]
a = 1.95 - 0.05n ...eqn.(3) \\[5ex]
\text{Substitute eqn.(3) into eqn.(2)} \\[3ex]
n(1.95 - 0.05n) + 1.9n = 46.5 \\[3ex]
1.95n - 0.05n^2 + 1.9n = 46.5 \\[3ex]
-0.05n^2 + 3.85n - 46.5 = 0 \\[3ex]
\text{Multiply both sides by } -100 \\[3ex]
-100(-0.05n^2 + 3.85n - 46.5) = -100(0) \\[3ex]
5n^2 - 385n + 4650 = 0 \\[3ex]
\text{Divide both sides by 5} \\[3ex]
n^2 - 77n + 930 = 0 \\[3ex]
(n - 15)(n - 62) = 0 \\[3ex]
n - 15 = 0 \;\;\;OR\;\;\; n - 62 = 0 \\[3ex]
n = 15 \;\;\;OR\;\;\; n = 62 \\[3ex]
\text{Substitute these values of n in eqn.(3)} \\[3ex]
For\;\;n = 15 \\[3ex]
a = 1.95 - 0.05(15) \\[3ex]
a = 1.95 - 0.75 \\[3ex]
a = 1.2 \\[5ex]
For\;\;n = 62 \\[3ex]
a = 1.95 - 0.05(62) \\[3ex]
a = 1.95 - 3.1 \\[3ex]
a = -1.15 \\[3ex]
$
The height of the shortest student, a cannot be a negative value
So, we have to discard that value.
This also implies that the corresponding value of n will be discarded.
Therefore:
(a.) The number of students in the class is 15
(b.) The height of the shortest student is 1.2 m
(27.) (a.) How many numbers between 75 and 500 are divisible by 7?
(b.) The 8th term of an Arithmetic Progression (A.P.) is 5 times the third term while the 7th term is 9
greater than the 4th term.
Write the first five terms of the A.P.
How many numbers between $75$ and $500$ are divisible by $7$?
Between $75$ and $500$ typically means that $75$ and $500$ are not included
Even if they are, both numbers are NOT divisible by $7$
So, we need to find the number immediately greater than $75$ that is divisible by $7$
That number is $77$.
This implies that $77$ is the first term.
Also, we need to find the number immediately less than $500$ that is divisible by $7$
That number is $497$.
This implies that $497$ is the last term.
Because the numbers are divisible by $7$, the common difference is $7$
The count of the numbers divisible by $7$ is the number of terms
$
(a.) \\[3ex]
a = 77 \\[3ex]
AP_n = 497 \\[3ex]
d = 7 \\[3ex]
n = ? \\[3ex]
AP_n = a + d(n - 1) \\[3ex]
497 = 77 + 7(n - 1) \\[3ex]
497 - 77 = 7(n - 1) \\[3ex]
420 = 7(n - 1) \\[3ex]
7(n - 1) = 420 \\[3ex]
n - 1 = \dfrac{420}{7} \\[5ex]
n - 1 = 60 \\[3ex]
n = 60 + 1 \\[3ex]
n = 61 \\[3ex]
$
There are 61 numbers between 75 and 500 that are divisible by 7
$
(b.) \\[3ex]
AP_n = a + d(n - 1) \\[3ex]
AP_8 = a + 7d \\[3ex]
AP_3 = a + 2d \\[3ex]
AP_7 = a + 6d \\[3ex]
AP_4 = a + 3d \\[3ex]
AP_8 = 5 * AP_3 \\[3ex]
a + 7d = 5(a + 2d) \\[3ex]
a + 7d = 5a + 10d \\[3ex]
5a + 10d = a + 7d \\[3ex]
5a - a + 10d - 7d = 0 \\[3ex]
4a + 3d = 0 ...eqn.(1) \\[3ex]
Also: \\[3ex]
AP_7 = 9 + AP_4 \\[3ex]
a + 6d = 9 + (a + 3d) \\[3ex]
a + 6d = 9 + a + 3d \\[3ex]
a - a + 6d - 3d = 9 \\[3ex]
3d = 9 \\[3ex]
d = \dfrac{9}{3} \\[5ex]
d = 3 \\[3ex]
Substitute \:\:d = 3\:\:into\:\:eqn.(1) \\[3ex]
4a + 3d = 0 \\[3ex]
4a = -3d \\[3ex]
4a = -3(3) \\[3ex]
4a = -9 \\[3ex]
a = -\dfrac{9}{4} \\[5ex]
AP_2 = a + d = -\dfrac{9}{4} + 3 = -\dfrac{9}{4} + \dfrac{12}{4} = \dfrac{-9 + 12}{4} = \dfrac{3}{4} \\[5ex]
AP_3 = a + 2d = AP_2 + d = \dfrac{3}{4} + 3 = \dfrac{3}{4} + \dfrac{12}{4} = \dfrac{3 + 12}{4} =
\dfrac{15}{4} \\[5ex]
AP_4 = a + 3d = AP_3 + d = \dfrac{15}{4} + 3 = \dfrac{15}{4} + \dfrac{12}{4} = \dfrac{15 + 12}{4} =
\dfrac{27}{4}
\\[5ex]
AP_5 = a + 2d = AP_4 + d = \dfrac{27}{4} + 3 = \dfrac{27}{4} + \dfrac{12}{4} = \dfrac{27 + 12}{4} =
\dfrac{39}{4}
\\[5ex]
AP:\:\: -\dfrac{9}{4}, \dfrac{3}{4}, \dfrac{15}{4}, \dfrac{27}{4}, \dfrac{39}{4}
$
