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3 years ago

Find the next three terms of the sequence. Then write a rule for the sequence.

1 answer:

8 0

We notice it diviides by 3 each time or multiplies by 1/3
common ratio is 1/3
first term is 648

so next 3 terms are 8, 8/3, and 8/9

rule is an=648(1/3)^(n-1)

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Any 10th grader solve it <br>for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

5 0

3 years ago

HELP ASAP GIVING BRANLIST 100 POINTS!!!

Sedbober [7]

Rate of change between 1 and 3: find the y values for 1 and 3:

Rate of change = (6-1) / (3-1) = 5/2

Rate of change from 2 and 4:

= (10-3) / (4-2) = 7/2

The rate from 2 to 4 is larger because 7/2 is larger than 5/2

6 0

3 years ago

Read 2 more answers

Which of the following correctly describes a concave polygon?

enyata [817]

The answer to this one is A

7 0

3 years ago

Juan takes 15 hours to do a job that his father can do in 6 hours. If juan works for 4.5 hours before his father joins him, how

ad-work [718]

Work it exactly like the other one with Wilda and Karla.

Juan takes 15 hours to do a job ... he does 1/15 of it in an hour.
His father can do it in 6 hours ... he does 1/6 of it in an hour.

Juan works alone for 4.5 hours. In that time, he does 4.5/15 = 0.3
of the job. When his father joins him, there's only 0.7 of it to finish.

Working together, they do (1/15 + 1/6) = (2/30 + 5/30) = 7/30 of the job
each hour.

How many times do they have to do 7/30 of the job in order to finish
the 7/10 of it that remains ? That will be the number of hours they need.

(7/10) / (7/30) = 7/10 x 30/7 = <em>3 hours</em>.

Then they can knock off while it's still daylight, and go for a swim.

3 0

3 years ago

Which number line have a point that is equal to point p? Check all that apply

leva [86]

Answer:

a,c

Step-by-step explanation:

my brain is very large

6 0

3 years ago