The 3rd term of a harmonic progression is 15 and the 9th term is 6. Find the 11th term.
Answer:
5
6
7
8
In a harmonic progression, reciprocals form an arithmetic progression. Let $b_n=1/a_n$. Given: $b_3=1/15$, $b_9=1/6$ $d=\frac{1/6-1/15}{9-3}=\frac{1}{60}$ $b_{11}=b_9+2d=\frac{1}{6}+\frac{1}{30}=\frac{1}{5}$ $\boxed{a_{11}=5}$
The 3rd term of a harmonic progression is 15 and the 9th term is 6. Find the 11th term of the progression.
5
4
6
3
In an HP the reciprocals form an AP: $\frac{1}{15}$ is the 3rd term, $\frac{1}{6}$ is the 9th. $d = \frac{\frac{1}{6} - \frac{1}{15}}{9 - 3} = \frac{1/10}{6} = \frac{1}{60}$ $a_1 = \frac{1}{15} - 2\left(\frac{1}{60}\right) = \frac{2}{60}$ $a_{11} = \frac{2}{60} + 10\left(\frac{1}{60}\right) = \frac{12}{60} = \frac{1}{5}$ HP term $= \frac{1}{1/5}$ $\boxed{= 5}$
Two trains A & B are moving on parallel tracks in the same direction and at different speeds. Train B, 180 m long, is moving at 120 kph. Train A, 120 m long, is behind train B and moving at 180 kph. At a certain time, the clear distance between the trains is 550 m.
Find the distance travelled by train B when the front end of train A caught up with its tail end?
1650 m
1350 m
1100 m
1850 m
Find the distance travelled by train A when its front caught up with the tail end of train B?
2850 m
1100 m
1350 m
1650 m
How long will it take for train A to completely overtake train B?
51 s
55 s
48 s
39 s
Part 1.
Relative speed $=180-120=60$ kph $=16.667$ m/s. For the front of train A to catch the tail of train B, it must close the 550 m clear distance. $t=\frac{550}{16.667}=33$ s. Train B speed $=120$ kph $=33.333$ m/s, so distance travelled by B is $33.333(33)=1100$ m. $\boxed{1100\text{ m}}$
Part 2.
The closing time for train A's front to catch train B's tail is $t=\frac{550}{16.667}=33$ s. Train A speed $=180$ kph $=50$ m/s. Distance travelled by A $=50(33)=1650$ m. $\boxed{1650\text{ m}}$
Part 3.
For train A to completely overtake train B, it must close the clear distance plus both train lengths: $550+180+120=850$ m. Relative speed $=60$ kph $=16.667$ m/s. $t=\frac{850}{16.667}=51$ s. $\boxed{51\text{ s}}$
There are seven arithmetic means between 3 and 35. Find the sum of all the terms.
169
171
167
173
Seven arithmetic means between 3 and 35 create a sequence of 9 terms. Sum of an arithmetic sequence: $S_n=\frac{n}{2}(a_1+a_n)$. $S_9=\frac{9}{2}(3+35)=171$ $\boxed{171}$
If $a$, $b$, 27 form a geometric progression and 6, $a$, $b$ form an arithmetic progression, what is the value of $b$?
12
18
16
20
Since $a,b,27$ are in geometric progression, $b^2=27a$. Since $6,a,b$ are in arithmetic progression, $2a=6+b$, so $a=\frac{b+6}{2}$. Substitute: $b^2=27\left(\frac{b+6}{2}\right)$ $2b^2-27b-162=0$ $b=18$ is the positive solution. $\boxed{18}$
Find the sum of 12 geometric means between 3 and 24,576.
49,149
24,570
12,285
36,872
With 12 geometric means between 3 and 24,576, there are 14 terms total. $3r^{13}=24576 \Rightarrow r^{13}=8192=2^{13}$, so $r=2$. Sum of all 14 terms: $S_{14}=3\frac{2^{14}-1}{2-1}=49149$. Subtract the endpoints: $49149-3-24576=24570$. $\boxed{24,570}$
The 8th term of a progression whose 12th, 13th, and 14th terms are 79.5, 86, and 92.5, respectively.
66.5
60
53.5
47
The tenth term of the progression 18, 54, 162,...
13122
1062882
118098
354294
The arithmetic mean of two numbers whose geometric mean is 4 and harmonic mean is 1/8.
128
144
98
64
Part 1.
The 12th, 13th, and 14th terms are arithmetic with common difference $86-79.5=6.5$. The 8th term is 4 terms before the 12th: $a_8=79.5-4(6.5)=53.5$ $\boxed{53.5}$
Part 2.
The sequence is geometric with $a_1=18$ and $r=3$. $a_{10}=18(3^{9})=354294$ $\boxed{354294}$
Part 3.
Let the two numbers be $a$ and $b$. Geometric mean 4 gives $ab=4^2=16$. Harmonic mean $H=\frac{2ab}{a+b}$. Since arithmetic mean $A=\frac{a+b}{2}$, $H=\frac{ab}{A}$. $\frac{1}{8}=\frac{16}{A} \Rightarrow A=128$ $\boxed{128}$
The 3rd, 11th, and 19th term of a progression is 33, 85, and 137, respectively. What is the 35th term?
228
234.5
247.5
241
The harmonic mean of two numbers is 16 and their geometric mean is 12. What is their arithmetic mean?
8
9
13
15
What is the sum of all the even integers between 10 and 500?
62730
62520
62220
63230
Part 1.
The 3rd, 11th, and 19th terms are 8 term-spaces apart. Their values increase by $85-33=52$, so the common difference is $d=\frac{52}{8}=6.5$. From the 19th term to the 35th term is 16 spaces: $a_{35}=137+16(6.5)=241$ $\boxed{241}$
Part 2.
Let the numbers be $a$ and $b$. The geometric mean is 12, so $ab=12^2=144$. The harmonic mean is $H=\frac{2ab}{a+b}$. Since the arithmetic mean is $A=\frac{a+b}{2}$, $H=\frac{ab}{A}$. $16=\frac{144}{A} \Rightarrow A=9$ $\boxed{9}$
Part 3.
The even integers strictly between 10 and 500 are 12, 14, ..., 498. Number of terms: $n=\frac{498-12}{2}+1=244$. Sum: $S=\frac{244}{2}(12+498)=62220$. $\boxed{62220}$
Given the following vectors: $A = 2i + 4j$; $B = i + 5j$.
What is the cross product $A \times B$?
$2i - 3j + 6k$
$4i + 2j - 7k$
$0i + 0j + 6k$
$8i + 0j + 0k$
What is the dot product $A \bullet B$?
$11$
$22$
$33$
$44$
What is the angle between $A$ and $B$?
$15.26^\circ$
$18.32^\circ$
$14.21^\circ$
$12.47^\circ$
Part 1.
Write the 2D vectors as 3D vectors with zero $k$ components: $A=2i+4j+0k$, $B=i+5j+0k$. $A\times B=(0)i+(0)j+(2(5)-4(1))k=6k$. $\boxed{0i+0j+6k}$
Part 2.
$A\bullet B=(2)(1)+(4)(5)=2+20=22$. $\boxed{22}$
Part 3.
Use $A\bullet B=|A||B|\cos\theta$. $A\bullet B=22$, $|A|=\sqrt{2^2+4^2}=\sqrt{20}$, and $|B|=\sqrt{1^2+5^2}=\sqrt{26}$. $\cos\theta=\frac{22}{\sqrt{20}\sqrt{26}}$, so $\theta=15.26^\circ$. $\boxed{15.26^\circ}$
Use $A\bullet B=|A||B|\cos\theta$. $A\bullet B=-11$, $|A|=\sqrt{6^2+(-3)^2+2^2}=7$, and $|B|=\sqrt{(-2)^2+1^2+2^2}=3$. $\cos\theta=\frac{-11}{7(3)}=-\frac{11}{21}$, so $\theta=121.58^\circ$. $\boxed{121.58^\circ}$
Part 4.
The scalar projection of $B$ on $A$ is $\frac{B\bullet A}{|A|}$. $B\bullet A=-11$ and $|A|=7$. Scalar projection $=-\frac{11}{7}$. $\boxed{-\frac{11}{7}}$
Part 5.
The vector projection of $B$ on $A$ is $\frac{B\bullet A}{|A|^2}A$. $B\bullet A=-11$ and $|A|^2=49$. $\text{proj}_A B=\frac{-11}{49}(6i-3j+2k)$ $=-1.35i+0.674j-0.45k$. $\boxed{-1.35i+0.674j-0.45k}$
Given the following vectors: $A = (xy)i + (2yz)j + (3xz)k$, $B = (yz)i + (2zx)j + (3xy)k$.
Determine the magnitude of vector $A$ at $(3, 2, -4)$.
$28.17$
$39.85$
$43.91$
$49.22$
Determine the magnitude of the resultant vector at $(3, 2, -4)$.
$43.91$
$49.22$
$39.85$
$28.17$
Determine the angle between the vectors at $(3, 2, -4)$.
$69.32^\circ$
$87.65^\circ$
$104.61^\circ$
$134.72^\circ$
Part 1.
At $(x,y,z)=(3,2,-4)$: $A=(xy)i+(2yz)j+(3xz)k=6i-16j-36k$. $|A|=\sqrt{6^2+(-16)^2+(-36)^2}=39.85$. $\boxed{39.85}$
Part 2.
At $(3,2,-4)$, $A=6i-16j-36k$. $B=(yz)i+(2zx)j+(3xy)k=-8i-24j+18k$. Resultant: $A+B=-2i-40j-18k$. $|A+B|=\sqrt{(-2)^2+(-40)^2+(-18)^2}=43.91$. $\boxed{43.91}$
Part 3.
At $(3,2,-4)$, $A=6i-16j-36k$ and $B=-8i-24j+18k$. $A\bullet B=6(-8)+(-16)(-24)+(-36)(18)=-312$. $|A|=39.85$ and $|B|=\sqrt{(-8)^2+(-24)^2+18^2}=31.05$. $\cos\theta=\frac{-312}{39.85(31.05)}$, so $\theta=104.61^\circ$. $\boxed{104.61^\circ}$
