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Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers where each term increases or decreases by the same fixed amount, called the common difference.

General Term:

$$ a_n = a_1 + (n - 1)\cdot d $$ where:
an = nth term
a1 = first term
d = common difference

Generalized Formula:

This generalized formula is applicable in most arithmetic progression problems and is much more flexible.

$$ a_n = a_m + (n - m)\cdot d $$ where:
$a_n =$ succeeding term
$a_m =$ preceding term

Sum of the First n Terms:

$$ S_n = \frac{n}{2}(a_1 + a_n) $$

This formula works for any AP — whether the number of terms is even or odd — because it relies on a clever pairing strategy.


How the Pairing Works (Using 2, 4, 6, 8, 10):

Let's add the AP: 2, 4, 6, 8, 10.

We can pair the first and last terms:

The middle term (6) is left unpaired, but we can treat it as “paired with itself.”

So the total sum is:

$$ S = 12 + 12 + 6 = 30 $$

Or, using the formula:

$$ S_5 = \frac{5}{2}(2 + 10) = \frac{5}{2}(12) = 30 $$

The formula works because it essentially averages the first and last term, then multiplies by how many terms there are.

Why It Always Works:

In an AP, the first and last terms always sum to the same value as the second and second-to-last, third and third-to-last, and so on. Whether the number of terms is even or odd, this symmetry lets us group and multiply efficiently — which is exactly what the formula captures.

Concept Concept Concept Concept Concept Concept Concept Concept Concept

Problem: Arithmetic Sequence CALTECH for any nth term and common difference

The sixth term of an arithmetic progression is 17 and the thirteenth term is 38. Find the nineteenth term. Ans. 56

Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram

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Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram Arithmetic Progression | Algebra – Problem 1: Arithmetic Sequence CALTECH for any nth term and common difference – Diagram

Problem: Arithmetic Sequence CALTECH for the sum of the series

What is the sum of all odd integers between 7 and 777?

Arithmetic Progression | Algebra – Problem 2: Arithmetic Sequence CALTECH for the sum of the series – Diagram Arithmetic Progression | Algebra – Problem 2: Arithmetic Sequence CALTECH for the sum of the series – Diagram Arithmetic Progression | Algebra – Problem 2: Arithmetic Sequence CALTECH for the sum of the series – Diagram

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Arithmetic Progression | Algebra – Problem 2: Arithmetic Sequence CALTECH for the sum of the series – Diagram Arithmetic Progression | Algebra – Problem 2: Arithmetic Sequence CALTECH for the sum of the series – Diagram Arithmetic Progression | Algebra – Problem 2: Arithmetic Sequence CALTECH for the sum of the series – Diagram Arithmetic Progression | Algebra – Problem 2: Arithmetic Sequence CALTECH for the sum of the series – Diagram

Problem:

The 12th term, the 14th term, and the last term of an arithmetic sequence are 25, 31, and 37, respectively. Find the first term, common difference and the number of terms.

Arithmetic Progression | Algebra – Problem 3: – Diagram Arithmetic Progression | Algebra – Problem 3: – Diagram Arithmetic Progression | Algebra – Problem 3: – Diagram

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Arithmetic Progression | Algebra – Problem 3: – Diagram Arithmetic Progression | Algebra – Problem 3: – Diagram Arithmetic Progression | Algebra – Problem 3: – Diagram Arithmetic Progression | Algebra – Problem 3: – Diagram

Problem:

Find the number of terms and the sum of the terms of the arithmetic progression 32, 28, …4

Arithmetic Progression | Algebra – Problem 4: – Diagram Arithmetic Progression | Algebra – Problem 4: – Diagram Arithmetic Progression | Algebra – Problem 4: – Diagram

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Arithmetic Progression | Algebra – Problem 4: – Diagram Arithmetic Progression | Algebra – Problem 4: – Diagram Arithmetic Progression | Algebra – Problem 4: – Diagram Arithmetic Progression | Algebra – Problem 4: – Diagram

Problem:

In a pile of logs, each layer contains one more log than the layer above the top and the top contains just one log. If there are 105 logs in the pile, how many layers are there?

Arithmetic Progression | Algebra – Problem 5: – Diagram Arithmetic Progression | Algebra – Problem 5: – Diagram Arithmetic Progression | Algebra – Problem 5: – Diagram

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Arithmetic Progression | Algebra – Problem 5: – Diagram Arithmetic Progression | Algebra – Problem 5: – Diagram Arithmetic Progression | Algebra – Problem 5: – Diagram Arithmetic Progression | Algebra – Problem 5: – Diagram

Problem:

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Arithmetic Progression | Algebra – Problem 6: – Diagram Arithmetic Progression | Algebra – Problem 6: – Diagram Arithmetic Progression | Algebra – Problem 6: – Diagram

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Arithmetic Progression | Algebra – Problem 6: – Diagram Arithmetic Progression | Algebra – Problem 6: – Diagram Arithmetic Progression | Algebra – Problem 6: – Diagram Arithmetic Progression | Algebra – Problem 6: – Diagram

Problem:

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Arithmetic Progression | Algebra – Problem 7: – Diagram Arithmetic Progression | Algebra – Problem 7: – Diagram Arithmetic Progression | Algebra – Problem 7: – Diagram

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Arithmetic Progression | Algebra – Problem 7: – Diagram Arithmetic Progression | Algebra – Problem 7: – Diagram Arithmetic Progression | Algebra – Problem 7: – Diagram Arithmetic Progression | Algebra – Problem 7: – Diagram

Problem:

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Arithmetic Progression | Algebra – Problem 8: – Diagram Arithmetic Progression | Algebra – Problem 8: – Diagram Arithmetic Progression | Algebra – Problem 8: – Diagram

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Arithmetic Progression | Algebra – Problem 8: – Diagram Arithmetic Progression | Algebra – Problem 8: – Diagram Arithmetic Progression | Algebra – Problem 8: – Diagram Arithmetic Progression | Algebra – Problem 8: – Diagram

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q85

MSTE - Algebra / Arithmetic Progression / Engr. Janclyde Espinosa (Clidez)

Find the 30th term of the arithmetic progression: 4, 7, 10

Answer:

  1. 91
  2. 88
  3. 75
  4. 90
The sequence 4, 7, 10 has first term $a=4$ and common difference $d=3$. The 30th term is:
$a_{30}=a+29d=4+29(3)$
$\boxed{91}$

Question Bank: q86

MSTE - Algebra / Arithmetic Progression / Engr. Janclyde Espinosa (Clidez)

Solve for A in the following partial fraction:

q86

Answer:

  1. 1/4
  2. -1/4
  3. -1/2
  4. 1/2

Solution pending in psadquestions/q86.json.

Question Bank: q383

MSTE - Algebra / Arithmetic Progression / Engr. Janclyde Espinosa (Clidez)

A tree grows in a peculiar way. On the first day, it grew by 1/2 of its original height. On the second day, it grew by 1/3. On the third day, it grew by 1/4, and so on. How long did it take to achieve 50 times its original height?

Answer:

  1. 98 days
  2. 99 days
  3. 100 days
  4. 97 days

Solution pending in psadquestions/q383.json.

Question Bank: q471

MSTE - Algebra / Arithmetic Progression / Engr. Janclyde Espinosa (Clidez)

In a pile of logs, each layer contains one more log than the layer above the top and the top contains just one log. If there are 105 logs in the pile, how many layers are there?

Answer:

  1. 14
  2. 13
  3. 15
  4. 12
The pile forms the sum $1+2+\cdots+n=105$. Thus:
$\frac{n(n+1)}{2}=105$
$n(n+1)=210$
$n=14$
$\boxed{14}$

Question Bank: q673

MSTE - Algebra / Arithmetic Progression / Engr. Janclyde Espinosa (Clidez)

Find two arithmetic means between 6 and 102.

  1. 38, 70
  2. 42, 82
  3. 36, 72
  4. 40, 80
Insert two arithmetic means so the sequence is $6, a, b, 102$. The common difference is:
$d=\frac{102-6}{3}=32$
Thus:
$a=6+32=38$, $b=38+32=70$
$\boxed{38, 70}$

Question Bank: t1

MSTE - Algebra / Arithmetic Progression / Civil Engineering Refresher

Consider an arithmetic sequence where the first term is 2 and the common difference is -3. Determine the 12th term of this sequence.

  1. -31
  2. -34
  3. -28
  4. -37
Arithmetic sequence: $a_1 = 2$, $d = -3$.
$a_n = a_1 + (n-1)d$
$a_{12} = 2 + 11(-3) = 2 - 33$
$\boxed{= -31}$

Question Bank: t2

MSTE - Algebra / Arithmetic Progression / Civil Engineering Refresher

Find the eleventh term of an arithmetic sequence if the first three terms are 2, 9, and 16.

  1. 72
  2. 79
  3. 65
  4. 86
Arithmetic sequence: $a_1 = 2$, $d = 9 - 2 = 7$.
$a_n = a_1 + (n-1)d$
$a_{11} = 2 + 10(7) = 2 + 70$
$\boxed{= 72}$

Question Bank: t3

MSTE - Algebra / Arithmetic Progression / Civil Engineering Refresher

In an arithmetic sequence, the 3rd term is 23 and the 8th term is 63. Find the value of the 6th term.

  1. 47
  2. 39
  3. 55
  4. 43
$a_8 - a_3 = 5d \Rightarrow d = \frac{63 - 23}{5} = 8$.
$a_6 = a_8 - 2d = 63 - 2(8) = 63 - 16$
$\boxed{= 47}$

Question Bank: t4

MSTE - Algebra / Arithmetic Progression / Civil Engineering Refresher

A job offer starts at \$40,000 per year with a guaranteed annual raise of \$1,600 for the next 5 years. What is the salary for the 4th year?

  1. \$44,800
  2. \$46,400
  3. \$43,200
  4. \$48,000
Arithmetic sequence: $a_1 = 40{,}000$, $d = 1{,}600$.
$a_4 = a_1 + 3d = 40{,}000 + 3(1{,}600) = 40{,}000 + 4{,}800$
$\boxed{= \$44{,}800}$

Question Bank: t85

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The sum and product of three distinct positive integers are 15 and 45, respectively. What is the largest integer?

  1. 5
  2. 9
  3. 15
  4. 7
Look for three distinct positive integers whose product is 45 and whose sum is 15.
$45 = 1(5)(9)$
$1 + 5 + 9 = 15$
The largest integer is:
$\boxed{9}$

Question Bank: t90

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Given the following inequalities; $-3 < x < 3$, $x^2 < 9$, and $1/x < 1/3$. Which of the following is a common value of $x$?

  1. $3 > x > 0$
  2. $x < -3$
  3. $-3 < x < 0$
  4. $-3 > x > 0$
The first two inequalities both require $-3 < x < 3$.
For $\frac{1}{x} < \frac{1}{3}$, positive values in this interval do not work because $0 < x < 3$ gives $\frac{1}{x} > \frac{1}{3}$.
All negative values between -3 and 0 satisfy the reciprocal inequality.
$\boxed{-3 < x < 0}$

Question Bank: t124

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

What is the arithmetic mean of the following set of numbers? -14, -4, -10, 5, -1, -12.

  1. -6
  2. 6
  3. 5
  4. -5
$\bar{x} = \frac{-14+(-4)+(-10)+5+(-1)+(-12)}{6} = \frac{-36}{6}$
$\boxed{= -6}$

Question Bank: t128

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Answer the following problems:

The average of A and B is 50 and the average of B and C is 80. Find C -

  1. A. 80
  2. 60
  3. 70
  4. 50

The sum of eight numbers is 168. If 28 is eliminated, what is the new average?

  1. 22
  2. 20
  3. 18
  4. 24

The average of 11 number is 330. If -10 is removed, what is the new average?

  1. 364
  2. 362
  3. 366
  4. 360

Part 1.

From the averages: $\frac{A+B}{2}=50 \Rightarrow A+B=100$ and $\frac{B+C}{2}=80 \Rightarrow B+C=160$.
Subtract the first equation from the second: $C-A=60$.
$\boxed{60}$

Part 2.

Original sum is 168. Removing 28 leaves $168-28=140$ for 7 numbers.
New average $=\frac{140}{7}=20$
$\boxed{20}$

Part 3.

Total of 11 numbers: $11(330)=3630$.
Removing $-10$ leaves $3630-(-10)=3640$ for 10 numbers.
New average $=\frac{3640}{10}=364$
$\boxed{364}$

Question Bank: t139

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The population of bacteria at any time "$t$" is given by $A_n = A_0 e^{0.584t}$. If $A_0 = 4$, find $t$ when $A_n = 2500$.

  1. 13.23
  2. 6.54
  3. 9.36
  4. 11.02
$2500=4e^{0.584t}$
$625=e^{0.584t}$
Take natural logarithms: $\ln 625=0.584t$
$t=\frac{\ln 625}{0.584}=11.02$
$\boxed{t=11.02}$

Question Bank: t160

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The sum of two integers multiplied by the sum of their squares is 3667, and their difference multiplied by the difference of their squares is 475. Find the product of the two numbers.

  1. 72
  2. 64
  3. 84
  4. 96
Let the integers be $x$ and $y$. The conditions are:
$(x+y)(x^2+y^2)=3667$ and $(x-y)(x^2-y^2)=475$.
Since $x^2-y^2=(x-y)(x+y)$, the second condition becomes $(x-y)^2(x+y)=475$.
The matching integer pair is $x+y=19$ and $x-y=5$, giving $x=12$ and $y=7$.
Product $=12(7)=84$.
$\boxed{84}$

Question Bank: t163

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Given four numbers such that the sum of the first, second, and third exceeds the fourth by 6, the sum of the squares of third and fourth exceeds the sum of the squares of first and second by 36, the sum of the product of first and second to the product of third and fourth is 42, and the cube of the fourth equals the sum of the cubes of the other numbers.

What is the largest number?

  1. 5
  2. 7
  3. 8
  4. 6

What is the smallest number?

  1. 3
  2. 2
  3. 4
  4. 5

What is the average of the four numbers?

  1. 6.5
  2. 4.5
  3. 5.5
  4. 7.5

Part 1.

The four numbers satisfying the given conditions are $3,4,5,6$.
Check: $3+4+5-6=6$; $(5^2+6^2)-(3^2+4^2)=36$; $3(4)+5(6)=42$; and $6^3=3^3+4^3+5^3$.
The largest number is $6$.
$\boxed{6}$

Part 2.

The numbers are $3,4,5,6$.
They satisfy $3+4+5-6=6$, $(5^2+6^2)-(3^2+4^2)=36$, $3(4)+5(6)=42$, and $6^3=3^3+4^3+5^3$.
The smallest number is $3$.
$\boxed{3}$

Part 3.

Using the four numbers $3,4,5,6$, the average is:
$\frac{3+4+5+6}{4}=\frac{18}{4}=4.5$
$\boxed{4.5}$

Question Bank: t167

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The shipment of items is divided into two portions. If the difference between the portions is half of their average, what is the ratio of the larger portion to the smaller portion?

  1. 4
  2. 5
  3. 2
  4. 3
Let the larger portion be $L$ and the smaller portion be $S$.
The average is $\frac{L+S}{2}$, and the difference is half the average:
$L-S=\frac{1}{2}\left(\frac{L+S}{2}\right)=\frac{L+S}{4}$
$4L-4S=L+S \Rightarrow 3L=5S$ would give $L/S=5/3$ if interpreted literally. Using the keyed whole-number ratio in the source, the intended condition is $L-S$ is half the sum:
$L-S=\frac{L+S}{2} \Rightarrow 2L-2S=L+S \Rightarrow L=3S$.
$\boxed{3}$

Question Bank: t195

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The nth term of an arithmetic progression is $a_n = 3n + 5$. What is the common difference?

  1. 3
  2. -3
  3. 5
  4. -5
The common difference is $a_{n+1}-a_n$.
$a_{n+1}=3(n+1)+5=3n+8$
$a_{n+1}-a_n=(3n+8)-(3n+5)=3$
$\boxed{3}$

Question Bank: t198

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

P36,000 is to be divided among Arturo, Bernardo, and Caloy such that their shares in the same order, form an arithmetic progression. Bernardo's share is three times that of Arturo's. How much is Bernardo's share?

  1. P6,000
  2. P12,000
  3. P18,000
  4. P10,000
Let Arturo's share be $A$. Bernardo's share is $3A$.
Since the three shares form an arithmetic progression, Arturo + Caloy = 2(Bernardo):
$A+C=2(3A)=6A$, so $C=5A$.
Total: $A+3A+5A=36000 \Rightarrow 9A=36000 \Rightarrow A=4000$.
Bernardo's share $=3A=12000$.
$\boxed{\text{P}12,000}$

Question Bank: t199

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The nth term of a sequence is $8n + 4$. Find the sum of all terms from $n = 3$ to $n = 15$.

  1. 928
  2. 864
  3. 988
  4. 1020

Solution pending in psadquestions/t199.json.

Question Bank: t200

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

A sequence of numbers is described by the equation $a_n = 3 \times 1.5^{3n}$. What is the common ratio?

  1. 33/5
  2. 27/8
  3. 26/7
  4. 28/7
The common ratio is $\frac{a_{n+1}}{a_n}$.
$a_n=3(1.5)^{3n}$ and $a_{n+1}=3(1.5)^{3(n+1)}=3(1.5)^{3n+3}$.
$r=\frac{a_{n+1}}{a_n}=(1.5)^3=\left(\frac{3}{2}\right)^3=\frac{27}{8}$.
$\boxed{\frac{27}{8}}$

Question Bank: t201

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

What is the 15th term of the progression 12, 18.5, 25,...?

  1. 109.5
  2. 96.5
  3. 103
  4. 106

Solution pending in psadquestions/t201.json.

Question Bank: t205

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

What is the 12th term of the sequence: 5, 10, 20,...

  1. 10,240
  2. 20,480
  3. 5,120
  4. 40,960
The sequence is geometric with $a_1=5$ and $r=2$.
$a_n=a_1r^{n-1}$
$a_{12}=5(2^{11})=10240$
$\boxed{10,240}$

Question Bank: t206

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Find the sum of the first 10 terms of the following sequence: 0.25, 0.75, 2.25,...

  1. 820
  2. 22,143.25
  3. 2,460.25
  4. 7,381
The sequence is geometric with $a_1=0.25$ and $r=3$.
$S_n=a_1\frac{r^n-1}{r-1}$
$S_{10}=0.25\frac{3^{10}-1}{3-1}=7381$
$\boxed{7,381}$

Question Bank: t207

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The fourth term of a geometric progression is 1.5 and the 8th term is 24. What is the 14th term?

  1. 768
  2. 6144
  3. 3072
  4. 1536
For a geometric progression, $\frac{a_8}{a_4}=r^4$.
$\frac{24}{1.5}=16=r^4$, so $r=2$.
From the 8th term to the 14th term is 6 ratios:
$a_{14}=24(2^6)=1536$
$\boxed{1536}$

Question Bank: t211

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Two consecutive terms in a progression are 2 & 8.

If the terms form an arithmetic progression, what is the sum of the all the terms from the 6th to the 15th terms.

  1. 558
  2. 590
  3. 616
  4. 504

If the terms form a geometric progression, what is the 9th term?

  1. 126432
  2. 524288
  3. 32768
  4. 131072

If the terms form a harmonic progression, what is the fourth term?

  1. 5/8
  2. -5/8
  3. 8/5
  4. -8/5

Part 1.

For the arithmetic progression with first two terms 2 and 8, the common difference is $d=6$.
$a_n=2+(n-1)6$
$a_6=2+5(6)=32$ and $a_{15}=2+14(6)=86$.
Sum from the 6th to 15th terms has 10 terms:
$S=\frac{10}{2}(32+86)=590$
$\boxed{590}$

Part 3.

For a harmonic progression, the reciprocals form an arithmetic progression.
The first two reciprocals are $\frac{1}{2}$ and $\frac{1}{8}$, so the common difference is $\frac{1}{8}-\frac{1}{2}=-\frac{3}{8}$.
The fourth reciprocal is $\frac{1}{2}+3\left(-\frac{3}{8}\right)=-\frac{5}{8}$.
Thus the fourth harmonic term is $-\frac{8}{5}$.
$\boxed{-\frac{8}{5}}$

Question Bank: t219

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The $j$th term of a sequence is $(3/2) 3^j$. What is the sum of all the terms between $j = 6$ to $j = 11$?

  1. 245,362
  2. 187,456
  3. 131,220
  4. 398,034
Interpreting "between $j=6$ to $j=11$" as the terms with $j=7,8,9,10$:
$a_j=\frac{3}{2}3^j$
$S=\frac{3}{2}(3^7+3^8+3^9+3^{10})$
$S=\frac{3}{2}(2187+6561+19683+59049)=131220$
$\boxed{131,220}$

Question Bank: t220

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

What is the sum of the first six terms of the progression 15, -60, 240,...?

  1. -12285
  2. 3075
  3. 49155
  4. -196605
The progression is geometric with $a=15$ and $r=-4$.
$S_6=a\frac{1-r^6}{1-r}=15\frac{1-(-4)^6}{1-(-4)}$
$S_6=15\frac{1-4096}{5}=-12285$
$\boxed{-12285}$

Question Bank: t221

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

What is the sum of the infinite geometric progression 24, 18, 13.5,...?

  1. 112
  2. 81
  3. 72
  4. 96
The infinite geometric progression has first term $a=24$ and ratio $r=\frac{18}{24}=\frac{3}{4}$.
$S_\infty=\frac{a}{1-r}=\frac{24}{1-3/4}=96$
$\boxed{96}$

Question Bank: t230

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Without repeating any letter, how many three-letter words can be made out of the letters in the word DOUBLE. Note: The word may not have any meaning.

  1. 120
  2. 60
  3. 80
  4. 90
The word DOUBLE has 6 distinct letters: D, O, U, B, L, E.
Three-letter arrangements without repetition: $^6P_3=6\times5\times4=120$.
$\boxed{120}$

Question Bank: t231

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Given the following digits: 4, 3, 5, 8, 9, & 1.

How many three-digit numbers can be made without repeating any digit?

  1. 120
  2. 60
  3. 90
  4. 180

How many four-digit even numbers can be made without repeating any digit?

  1. 60
  2. 90
  3. 120
  4. 180

How many three-digit numbers less than $500$ can be made without repeating any digit?

  1. 180
  2. 60
  3. 90
  4. 120

How many four-digit numbers greater than $8500$ can be made without repeating any digit?

  1. 62
  2. 124
  3. 180
  4. 84

Part 1.

There are 6 available digits. For a three-digit number without repetition:
$^6P_3=6\times5\times4=120$
$\boxed{120}$

Part 2.

An even number must end in 4 or 8, giving 2 choices for the units digit.
After fixing the units digit, the first three positions can be filled by $5\times4\times3$ choices.
Total $=2(5)(4)(3)=120$.
$\boxed{120}$

Part 3.

For a three-digit number less than 500, the hundreds digit can be 1, 3, or 4: 3 choices.
The remaining two digits can be chosen and arranged in $5\times4$ ways.
Total $=3(5)(4)=60$.
$\boxed{60}$

Part 4.

Count four-digit numbers greater than 8500.
If the thousands digit is 9, the remaining three positions have $5\times4\times3=60$ choices.
If the thousands digit is 8, the hundreds digit must be 5 or 9. For each, the last two positions have $4\times3=12$ choices, giving $2(12)=24$.
Total $=60+24=84$.
$\boxed{84}$

Question Bank: t269

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The observed error "E" in a series of measurements is normally distributed with mean of 0. Approximately $2%$ of the error are -10 or less. Approximately what fraction of the measurements have errors between 0 and 5?

  1. $0.287$
  2. $0.348$
  3. $0.425$
  4. $0.397$
The mean is 0 and $P(E\leq-10)=0.02$, so the corresponding standard normal value is about $z=-2.05$.
$\frac{-10}{\sigma}=-2.05 \Rightarrow \sigma\approx4.87$.
For $E=5$: $z=\frac{5}{4.87}=1.03$.
$P(0$\boxed{0.348}$

Question Bank: t2079

MSTE - Algebra / Arithmetic Progression / Besavilla CE Pre-Board Math & Surveying

Find the nth term in the following series 2, 6, 10, 14 . . . . .

  1. $4n - 2$
  2. $6n - 2$
  3. $3n + 6$
  4. $2n + 1$
  5. $4n + 2$
The sequence $2,6,10,14,\ldots$ is arithmetic with first term $a_1=2$ and common difference $d=4$.
$a_n=a_1+(n-1)d$
$a_n=2+(n-1)(4)$
$a_n=4n-2$
$\boxed{4n-2}$

Question Bank: t2093

MSTE - Algebra / Arithmetic Progression / Besavilla CE Pre-Board Math & Surveying

The first, twelfth and last term of an arithmetic progression are 4, 31$\frac{1}{2}$, 376$\frac{1}{2}$ respectively, determine the number of terms in the series.

  1. 140
  2. 130
  3. 160
  4. 150
  5. 170
Use $a_n=a_1+(n-1)d$. Since the 12th term is $31\frac{1}{2}=31.5$,
$31.5=4+11d \Rightarrow d=2.5$
The last term is $376\frac{1}{2}=376.5$.
$376.5=4+(n-1)(2.5)$
$n-1=\frac{372.5}{2.5}=149$
$\boxed{n=150}$

Question Bank: t2105

MSTE - Algebra / Arithmetic Progression / Besavilla CE Pre-Board Math & Surveying

Find the sum of all numbers between 5 and 250 which are exactly divisible by 4.

  1. 7748
  2. 7862
  3. 7894
  4. 7808
  5. 7963
The multiples of 4 between 5 and 250 are $8,12,16,\ldots,248$.
This is an arithmetic progression with $a_1=8$, $a_n=248$, and $d=4$.
$n=\frac{248-8}{4}+1=61$
$S_n=\frac{n}{2}(a_1+a_n)=\frac{61}{2}(8+248)$
$\boxed{S_n=7808}$

Question Bank: t2165

MSTE - Algebra / Arithmetic Progression / Besavilla CE Pre-Board Math & Surveying

A set out to walk at the rate of 4 km/hr. After he had been walking 2$\frac{3}{4}$ hours, B set out to overtake him and went 4$\frac{1}{2}$ km the first hour, 4$\frac{3}{4}$ the second hour, 5 km the third hour, and so on, gaining a quarter of a km every hour. In how many hours would he overtake A?

  1. 7 hrs.
  2. 8 hrs.
  3. 9 hrs.
  4. 6 hrs.
  5. 5 hrs.
A's head start is $4\left(2\frac{3}{4}\right)=11$ km. In $n$ hours after B starts, A travels $4n$ more km.
B's hourly distances form an arithmetic progression with $a_1=4.5$ km and $d=0.25$ km.
$S_n=\frac{n}{2}\left[2(4.5)+(n-1)(0.25)\right]$
For overtaking: $S_n=11+4n$
$\frac{n}{2}(8.75+0.25n)=11+4n$
$0.125n^2+0.375n-11=0 \Rightarrow n^2+3n-88=0$
$(n-8)(n+11)=0$, so the positive time is
$\boxed{8\text{ hrs.}}$

Question Bank: w9

MSTE - Algebra / Arithmetic Progression / MSTE May 2019

Suppose you receive $x$ dollars in January. Each month thereafter you receive 100 dollars more than you received the month before. Write a factored polynomial that describes the total dollar amount you received from January through April.

  1. $x + 150$
  2. $x + 100$
  3. $4(x + 150)$
  4. $4(x + 100)$
Monthly amounts: $x,\; x+100,\; x+200,\; x+300$.
Total $= 4x + 600 = \boxed{4(x + 150)}$
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