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 ${\sf T_n}$ is given by ;

${\boxed{\bf T_{n}=T_{1}+(n-1)d}}$

Where , <u>d</u> is the common difference

Now , here we are given with ;

${\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}$

We have to find the 2nd , 3rd and 4th term respectively ,

Now , by using the above formula , 5th term can be written as ;

${: \implies \quad \sf T_{1}+(5-1)d=T_{5}}$

Putting the values and transposing 1st term to RHS , we have ;

${: \implies \quad \sf 4d = -4-8}$

${: \implies \quad \sf d=-\dfrac{12}{4}}$

${: \implies \quad \sf d=-3}$

Now , as we got the common difference , so we can find out the missing terms now ;

${: \implies \quad \sf T_{2}= T_{1}+(2-1)d}$

${: \implies \quad \sf T_{2}= 8 +d}$

${: \implies \quad \sf T_{2}= 8-3}$

${: \implies \quad \bf \therefore \: T_{2}= 5}$

Now

${: \implies \quad \sf T_{3}= T_{1}+(3-1)d}$

${: \implies \quad \sf T_{3}= 8 +2d}$

${: \implies \quad \sf T_{3}= 8-6}$

${: \implies \quad \bf \therefore \: T_{3}= 2}$

Also ,

${: \implies \quad \sf T_{4}= T_{1}+(4-1)d}$

${: \implies \quad \sf T_{4}= 8 +3d}$

${: \implies \quad \sf T_{4}= 8-9}$

${: \implies \quad \bf \therefore \: T_{4}= -1}$

Now , The given table can be written as ;

${\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}$

Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;

brainly.com/question/26750175

8 0

3 years ago

I need help on this problem!!! please help!!!

MA_775_DIABLO [31]

(0.1x-22)+(0.3x-54)=90

First combine like terms

0.4x - 76 = 90

Now add 76 on both sides

0.4x = 166

Divide by 0.4 on each side to isolate x

X= 415

7 0

3 years ago