Question

What is the 42 term where a1=-12 and a27=66

Answer 1

Answer:

111

Step-by-step explanation:

a1 = -12

a27 = 66

Now using the formula  an = a1+(n-1)d we will find the value of d

here n = 27

a1 = -12

a27 = 66

Now substitute the values in the formula:

a27 = -12+(27-1)d

66= -12+(26)d

66 = -12+26 * d

66+12 = 26d

78 = 26d

now divide both the sides by 26

78/26= 26d/26

3 = d

Now put all the values in the formula to find the 42 term

an = a1+(n-1)d

a42 = -12 +(42-1)*3

a42 = -12+41 *3

a42 = -12+123

a42 = 111

Therefore 42 term is 111....

Answer 2

Answer:

Assuming it is arithmetic, the 42nd term is 111.

Assuming it is geometric, the conclusion says it isn't geometric.

Step-by-step explanation:

Let's assume arithmetic first.

Arithmetic sequences are linear. They go up or down by the same number over and over.  This is called the common difference.

We are giving two points on our line (1,-12) and (27,66).

Let's find the point-slope form of this line.

To do this I will need the slope.  The slope is the change of y over the change of x.

So I'm going to line up the points and subtract vertically, then put 2nd difference over 1st difference.

(1  , -12)

-(27,66)

------------

-26   -78

The slope is -78/-26=78/26=3.  The slope is also the common difference.

I'm going to use point $(x_1,y_1)=(1,-12)$ and $m=3$ in the point-slope form of a line:

$y-y_1=m(x-x_1)$

$y-(-12)=3(x-1)$

Distribute:

$y+12=3x-3$

Subtract 12 on both sides:

$y=3x-3-12$

$y=3x-15$

So we want to know what y is when x=42.

$y=3(42)-15$

$y=126-15$

$y=111$

So $a_{42}=111$ since the explicit form for this arithmetic sequence is

$a_n=3n-15$

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Let's assume not the sequence is geometric. That means you can keep multiplying by the same number over and over to generate the terms given a term to start with.  That is called the common ratio.

The explicit form of a geometric sequence is $a_n=a_1 \cdot r^{n-1}$.

We are given $a_1=-12$

so this means we have

$a_n=-12 \cdot r^{n-1}$.

We just need to find r, the common ratio.

If we divide 27th term by 1st term we get:

$\frac{a_{27}}{a_1}=\frac{-12r^{27-1}}{-12r^{1-1}}=\frac{-12r^{26}}{-12}=r^{26}$

We are also given this ration should be equal to 66/-12.

So we have

$r^{26}=\frac{66}{-12}$.

$r^{26}=-5.5$

So the given sequence is not geometric because we have an even powered r equaling a negative number.