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Luba_88 [7]
3 years ago
13

For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of

the first 10 terms in the sequence, then T isA. Greater than 2B. Between 1 and 2C. Between 1/2 and 1D. Between 1/4 and 1/2E. Less than 1/4
Mathematics
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is (-1)^{k+1}(\frac{1}{2^{k}}).

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as (-1)^{k+1}(\frac{1}{2^{k}}).

where k is from 1 to 10.

By the given expression sequence will be \frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......

In this sequence first term "a" = \frac{1}{2}

and common ratio in each successive term to the previous term is 'r' = \frac{\frac{(-1)}{4}}{\frac{1}{2} }

r = -\frac{1}{2}

Since the sequence is infinite and the formula to calculate the sum is represented by

S=\frac{a}{1-r} [Here r is less than 1]

S=\frac{\frac{1}{2} }{1+\frac{1}{2}}

S=\frac{\frac{1}{2}}{\frac{3}{2} }

S = \frac{1}{3}

Now we are sure that the sum of infinite terms is \frac{1}{3}.

Therefore, sum of 10 terms will not exceed \frac{1}{3}

Now sum of first two terms = \frac{1}{2}-\frac{1}{4}=\frac{1}{4}

Now we are sure that sum of first 10 terms lie between \frac{1}{4} and \frac{1}{3}

Since \frac{1}{2}>\frac{1}{3}

Therefore, Sum of first 10 terms will lie between \frac{1}{4} and \frac{1}{2}.

Option D will be the answer.

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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! This needs to be answered ASAP!
andriy [413]

Answer:

it will take 4 hours( I'm not so sure)

Step-by-step explanation:

$25x $15

------- = ------------- + 75

1h. 1h

25x = 100

x =100/25

x = 4

8 0
3 years ago
The sum of two numbers is 64. One number is 8 more than the other. What are the two numbers?
lisabon 2012 [21]

Answer:

28 and 36.

Step-by-step explanation:

Alright, so we are given that the sum of two numbers is 64. Based on this, we know that we are going to be adding because the question said "the sum of two numbers is 64." Sum is the answer to an addition problem. Lets write an equation.

We need a variable, let x be the variable which is the value of the smaller number.

the equation is x+x+8=64

x= to the smaller number

x+8= to the number that is eight more than x or the smaller number.

Before we isolate for x, we need to simplify the expression by combining the like terms.

x plus x equals 2x.

Therefore, we have 2x+8=64.

Our goal is to isolate x, so we need to subtract 8 from both sides. 8 mius 8 is 0, and 64 mius 8 is 56.

Now, we have 2x=56.

We still need to isolate x by dividing both sides by 2. 2 divided by 2 is 1, so we are left with 1x or x. 56 divided by 2 is 28.

x=28.

Great, we found what x or the smaller value is.

Now, we need to find the larger number.

Substitute x as 28 into x+8 to find the number that is 8 more than the smaller number or x. You'll then get 36. If you add 36 and 28, you get 64. Also, 36 is 8 more than 28, so this answer is correct.

Hope this helps!

4 0
3 years ago
Pq is parallel to rs the length of rp is 6cm the length of pt is 18cm the length of qt is 21cm what is the length of sq
nata0808 [166]
If this is the picture the answer is 7cm 

8 0
3 years ago
Read 2 more answers
a jar contains 16 marbles of which 1 is blue 5 red are red and the rest are green what is the ratio of green marbles to all marb
zloy xaker [14]

Answer: 10:16

Step-by-step explanation:

16-5-1=10

8 0
3 years ago
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