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3 years ago

Determine the next three terms in the following sequence: 1, 8, 27, 64,, , ,...

Mathematics

2 answers:

Strike441 [17]3 years ago

8 0

Answer:

d. 125, 216, 343

Step-by-step explanation:

The given sequence is 1, 8, 27, 64,.......

Now, we can write

$1=1^3\\\\8=2^3\\\\27=3^3\\\\64=4^3$

Here, we can see that the first term is cube of 1, second term is cube of 2, third term is cube of 3 and so on.

Therefore, in order to find the 5th, 6th and 7th term of the sequence, we have to take cube of 5, 6 and 7 respectively.

Therefore, the next three terms of the sequence are

$5^3=125\\\\6^3=216\\\\7^3=729$

The correct option is d which is 125, 216, 343

Dovator [93]3 years ago

4 0

Hello,

Answer D
$x_{n} =n^3\\
 x_{1} =1^3=1\\
 x_{2} =2^3=8\\
 x_{3} =3^3=27\\
 x_{4} =4^3=64\\
 x_{5} =5^3=125\\
 x_{6} =6^3=216\\
 x_{7} =7^3=343\\
$

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A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

4 0

3 years ago

Solve for x.7x + 14 – 2x = −x + 12 – x − 19A.x = −3B.x = −1C.x = 1D.x = 3

lara31 [8.8K]

7x+14-2x=-x+12-x-19
collect like terms
7x-2x+X+X+19=12-14
7x+19+2
7x=-21
x-21/7
X=-3

4 0

3 years ago

If you guys want me to join y’all’s class in like zoom then tell me your class id and password.

sammy [17]

Answer:

You cant join mine. Mine has strict rules and c rap sign up with a email and stuff.

Step-by-step explanation:

6 0

3 years ago

A coin is tossed 5 times; find the probability of getting at least 1 tail. Would you consider this event likely to happen?

grin007 [14]

Answer: 0.96875

Step-by-step explanation:

Probability of obtaining at least one tail on a toss of coin 5 times ;

P(success) in a coin toss = 1/2 = 0.5

Using the binomial distribution formula :

P(x) = nCx * p^x * (1 - p)^(n-x)

P(at least one tail) = 1 - p(0 tails)

P( 0 tails) = 5C0 * 0.5^0 * 0.5^(5-0)

= 1 * 1 * 0.03125

= 0.03125

P(at leat 1 tail) = 1 - 0.03125

P(atleast 1 tail) = 0.96875

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3 years ago

I NEED HELP PLEASE! ASAP PLEASE! Solve the problem. m + 7 > 13 m >

a_sh-v [17]

Answer:

m > 6

Step-by-step explanation:

Move all terms not containing m to the right side of the inequality.

Inequality Form:

m > 6

Interval Notation:

(6, ∞)

Therefor, the answer would be:

m > 6

Hope this helps!

4 0

3 years ago

Determine the next three terms in the following sequence: 1, 8, 27, 64,____, ____, ____,...

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