Respuesta :
Answer:
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
The first three terms of the arithmetic sequence are 29/2, 13, 23/2.
Given
The value of [tex]\rm t_8[/tex] = 4 and [tex]\rm t_{12}[/tex] = −2.
Arithmetic sequence;
The difference between two consecutive terms is constant is called an arithmetic sequence.
The general form of the arithmetic sequence.
[tex]\rm t_n=t_1+(n-1)d[/tex]
Where, [tex]\rm t_1[/tex] is the first term and d is the common difference of the arithmetic sequence.
The 8th term of the sequence is;
[tex]\rm t_8=t_1+(8-1)d\\\\t_8=t_1+7d[/tex]
The 12th term of the sequence is;
[tex]\rm t_{12}=t_1+(12-1)d\\\t_{12} \\\\ t_{12}=t_1+11d[/tex]
[tex]\rm t_{12}=t_1+11d[/tex]
Subtract both the equations;
[tex]\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}[/tex]
The common difference between the terms is -3/2.
Then,
The first term of the sequence is;
[tex]\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}\\\\[/tex]
The first term of the sequence is 29/2.
Therefore,
The second term of the sequence is;
[tex]\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[/tex]
And a third of the sequence is;
[tex]\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}\\\\[/tex]
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.
https://brainly.com/question/9857007
