Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;
The n'th term of a Arithmetic Sequence let's say it be [tex]{\sf T_n}[/tex] is given by ;
- [tex]{\boxed{\bf T_{n}=T_{1}+(n-1)d}}[/tex]
Where , d is the common difference
Now , here we are given with ;
[tex]{\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}[/tex]
We have to find the 2nd , 3rd and 4th term respectively ,
Now , by using the above formula , 5th term can be written as ;
[tex]{: \implies \quad \sf T_{1}+(5-1)d=T_{5}}[/tex]
Putting the values and transposing 1st term to RHS , we have ;
[tex]{: \implies \quad \sf 4d = -4-8}[/tex]
[tex]{: \implies \quad \sf d=-\dfrac{12}{4}}[/tex]
[tex]{: \implies \quad \sf d=-3}[/tex]
Now , as we got the common difference , so we can find out the missing terms now ;
[tex]{: \implies \quad \sf T_{2}= T_{1}+(2-1)d}[/tex]
[tex]{: \implies \quad \sf T_{2}= 8 +d}[/tex]
[tex]{: \implies \quad \sf T_{2}= 8-3}[/tex]
[tex]{: \implies \quad \bf \therefore \: T_{2}= 5}[/tex]
Now
[tex]{: \implies \quad \sf T_{3}= T_{1}+(3-1)d}[/tex]
[tex]{: \implies \quad \sf T_{3}= 8 +2d}[/tex]
[tex]{: \implies \quad \sf T_{3}= 8-6}[/tex]
[tex]{: \implies \quad \bf \therefore \: T_{3}= 2}[/tex]
Also ,
[tex]{: \implies \quad \sf T_{4}= T_{1}+(4-1)d}[/tex]
[tex]{: \implies \quad \sf T_{4}= 8 +3d}[/tex]
[tex]{: \implies \quad \sf T_{4}= 8-9}[/tex]
[tex]{: \implies \quad \bf \therefore \: T_{4}= -1}[/tex]
Now , The given table can be written as ;
[tex]{\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}[/tex]
Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;
https://brainly.com/question/26750175
