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3 years ago

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Mathematics

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Burka [1]3 years ago

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Arithmetic sequences et sn} be an arithmetic sequence that starts with an initial index of 0. The initial term is 3 and the comm

navik [9.2K]

Answer:

(a) The value of $s_z$ is (z+1)(3-z).

(b) The next term in the sequence is -2.

Step-by-step explanation:

(a)

It is given that arithmetic sequence that starts with an initial index of 0.

The initial term is 3 and the common difference is -2.

$a_0=3$

$d=-2$

We need to find the value of $s_z$ .

$s_z=\sum_{n=0}^{n=z}(a+nd)$

where, a is initial term and d is common difference.

$s_z=\sum_{n=0}^{n=z}(3-2n)$

The sum of an arithmetic sequence with initial index 0 is

$s_n=\frac{n+1}{2}[2a+nd]$

where, a is initial term and d is common difference.

Substitute n=z, a=3 and d=-2 in the above formula.

$s_z=\frac{z+1}{2}[2(3)+z(-2)]$

$s_z=\frac{z+1}{2}[2(3-z)]$

$s_z=(z+1)(3-z)$

Therefore the value of $s_z$ is (z+1)(3-z).

(b)

The given arithmetic sequence is

7, 4, 1, ...

We need to find the term in the sequence.

In the given arithmetic sequence the first term is

$a=7$

The common difference of the sequence is

$d=a_2-a_1\Rightarrow 4-7=-3$

The first term is 7 and common difference is -3.

Add common difference in last given term, i.e., 1, to find the next term of the sequence.

$1+(-3)=1-3=-2$

Therefore the next term in the sequence is -2.

3 0

4 years ago

How many people (x) must attend the third show so that the average attendance per show is 3000? Choose the correct equation.

Pavel [41]

I think it’s aaaaaasss

4 0

3 years ago

Read 2 more answers

To solve 493x = 3432x+1, write each side of the equation in terms of base

sleet_krkn [62]

Answer

= 3.405 × 10⁻⁴

Explanation

If the question is correctly written, then it is very easy.

First, take the like terms on one side.

493x = 3432x+1

493x - 3432x = 1

-2939x = 1

Dividing both sides by -2939 to get the value of x.

x = 1/(-2939)

= - 0.0003405

= 3.405 × 10⁻⁴

3 0

3 years ago

Read 2 more answers

Members of the track team can run 400 min an average time of 64.6 seconds. The

vova2212 [387]

Answer:

T maximum=T average -7.8 seconds

T minimum=T average +7.8 seconds

Step-by-step explanation:

Calculation for the equation that can be

use to find the maximum and minimum times for the track team

Using this equation to find the maximum times for the track team

T maximum=T average -7.8 seconds

T maximum=64.6 seconds-7.8 seconds

Using this equation to find the minimum times for the the track team

T minimum=T average +7.8 seconds

T minimum=64.6 seconds +7.8 seconds

Therefore the equation for the maximum and minimum times for the track team are :

T maximum=T average -7.8 seconds

T minimum=T average +7.8 seconds

6 0

3 years ago

Any 10th grader solve it 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

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

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

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

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

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

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