Mathematics, 21.06.2019 23:00 Tyrant4life

If t8 = 4 and t12 = −2, find the first three terms of the arithmetic sequence.

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Answer from: coolman5999alt

The first three terms of the arithmetic sequence are 29/2, 13, 23/2.

Given

The value of $\rm t_8$ = 4 and $\rm t_{12}$ = −2.

Arithmetic sequence;

The difference between two consecutive terms is constant is called an arithmetic sequence.

The general form of the arithmetic sequence.

$\rm t_n=t_1+(n-1)d$

Where, $\rm t_1$ is the first term and d is the common difference of the arithmetic sequence.

The 8th term of the sequence is;

$\rm t_8=t_1+(8-1)d\\\\t_8=t_1+7d$

The 12th term of the sequence is;

$\rm t_{12}=t_1+(12-1)d\\\t_{12} \\\\ t_{12}=t_1+11d$

$\rm t_{12}=t_1+11d$

Subtract both the equations;

$\rm t_8-t_{12}=t_1+7d-(t_1+11d)\\\\t_8-t_{12}=t_1+7d-t_1-11d\\\\ t_8-t_{12}=-4d\\\\ (4-(-2))=-4d\\\\4+2=-4d\\\\-4d=6\\\\d=\dfrac{6}{-4}\\\\d=\dfrac{-3}{2}$

The common difference between the terms is -3/2.

Then,

The first term of the sequence is;

$\rm t_8=t_1+7d\\\\4=t_1+7(\dfrac{-3}{2})\\\\t_1=4+\dfrac{21}{2}\\\\t_1=\dfrac{8+21}{2}\\\\t_1=\dfrac{29}{2}\\\\$

The first term of the sequence is 29/2.

Therefore,

The second term of the sequence is;

$\rm t_n=t_1+(n-1)d\\\\t_2=\dfrac{29}{2}+(2-1)(\dfrac{-3}{2})\\\\t_2=\dfrac{29}{2}-\dfrac{3}{2}\\\\t_2=\dfrac{29-3}{2}\\\\t_2=\dfrac{26}{2}\\\\t_2=13$

And a third of the sequence is;

$\rm t_n=t_1+(n-1)d\\\\t_3=\dfrac{29}{2}+(3-1)(\dfrac{-3}{2})\\\\t_2=\dfrac{29}{2}-\dfrac{6}{2}\\\\t_2=\dfrac{29-6}{2}\\\\t_2=\dfrac{23}{2}\\\\$

Hence, the first three terms of the arithmetic sequence are 29/2, 13, 23/2.

To know more about arithmetic sequence click the link given below.

link

Answer from: robert7248

14.5, 13, 11.5

Step-by-step explanation:

The general term (an) of an arithmetic sequence with first term a1 and common difference d is ...

an = a1 + d(n-1)

Then the 8th term is ...

a8 = a1 + d(8-1)

and the 12th term is ...

a12 = a1 + d(12-1)

So, the difference between these terms is ...

a12 -a8 = (a1 +11d) -(a1 +7d) = 4d

= (-2-4) = -6 . . . . . substituting values for a12 and a8

Then the common difference is

... d = -6/4 = -3/2

Using this, we can find a1 from a8.

4 = a1 +7·(-3/2) = a1 - 10.5

14.5 = a1 . . . . . . . add 10.5 to both sides of the equation

This is the first term. The second is this value with the common difference added:

14.5 + (-1.5) = 13

The third term is this with the common difference added:

13 + (-1.5) = 11.5

In summary, the first three terms are ...

14.5, 13, 11.5

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If t8 = 4 and t12 = −2, find the first three terms of the arithmetic sequence....