All categories

3 years ago

STD 10

Mathematics

1 answer:

OLga [1]3 years ago

5 0

Answer:

False

Step-by-step explanation:

Given

$6 , 12 , 18 , 24 , .......... , 284$

Required

Determine if it is an AP or not

First, calculate the common difference (d)

$d = 24 - 18=6$

$d = 18 - 12=6$

$d = 12 - 6=6$

Next, is to determine if the last term is valid using:

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

So, we have:

$284 = 6 + (n - 1) * 6$

$284 = 6 + 6n - 6$

Collect like terms

$284 = 6 - 6+ 6n$

$284 = 6n$

Divide by 6

$284/6 = n$

$n = 284/6$

$n = 47.33$

<em>The value of n must be a positive whole number. Hence, it is not an AP</em>

You might be interested in

Which of the following rational functions is graphed below?

vampirchik [111]

Answer:

The correct option is D.

i.e.

$f\left(x\right)=\frac{1}{x\left(x+4\right)}$ is the correct option.

The correct graph is shown in attached figure.

Step-by-step explanation:

Considering the function

$f\left(x\right)=\frac{1}{x\left(x+4\right)}$

$\mathrm{Domain\:of\:}\:\frac{1}{x\left(x+4\right)}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x$

$\mathrm{Range\:of\:}\frac{1}{x\left(x+4\right)}:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:-\frac{1}{4}\quad \mathrm{or}\quad \:f\left(x\right)>0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-\frac{1}{4}]\cup \left(0,\:\infty \:\right)\end{bmatrix}$