Question

find a8 for the arithmatic sequence a1=6x-9, a2=5x+1

Answer 1

is this one problem or 2 separate problems?

Answer 2

one problem

Answer 3

Find common difference:

\(a_2 - a_1 = 5x + 1 - 6x - 9\)

\(d = -x - 8\)


Now \(a_8\) is given by:

\[a_8 = a_1 + (8-1)(d) \implies a_8 = a_1 + 7d\]

can you find \(a_8\) here ??

Answer 4

maybe let me try

Answer 5

Take your time..

Answer 6

6x-9+7(-x-8) is that how I start it

Answer 7

Distribute 7 to the brackets..

Answer 8

Yes you are right

Solve this and have faith on yourself.

Answer 9

6x-9-7x-56

Answer 10

Yes carry on..

Answer 11

-x-65

Answer 12

Well Done..

Answer 13

is that the answer

Answer 14

\(a_8 = -x - 65\)

Yeah that looks good to me.

Answer 15

the back of my book says the answer is 61 not 65

Answer 16

We should then check it once again..

Answer 17

Oh sorry I have mistaken above..

Answer 18

with what

Answer 19

\(a_2 - a_1 = 5x + 1 - 6x + 9\)

\(d = -x + 10\)

Now do the same..

Only value of d was wrong..

Answer 20

why did you put plus 9

Answer 21

\(a_8 = 6x - 9 + 7(-x + 10) \)

\(a_8 = 6x - 9 - 7x + 70\)

Answer 22

Do you know:

-(-) becomes + ??

Answer 23

its a plus 1 tho

Answer 24

Where ??

Answer 25

5x+1 and -6x-9

Answer 26

+ implies +1 no doubt in this..

Answer 27

Yes now subtract them..

Answer 28

-8

Answer 29

\(5x + 9 - (6x - 9)\) = ??

Answer 30

5x+1

Answer 31

How??

Answer 32

Oh sorry my mistake there..

Answer 33

\(5x + 1 - (6x - 9) = ??\)

Answer 34

-x-8

Answer 35

No be careful..

Answer 36

ohhh i see

Answer 37

the negative before parenthasis

Answer 38

Yes..

Answer 39

Getting ??

Answer 40

got it

Answer 41

can u help with another

Answer 42

Sure..

Answer 43

find s12 a2=6, d=10

Answer 44

what is s12 ???

12th term ??

Answer 45

ya

Answer 46

See, there is a general formula to find any term:

\[\large a_n = a + (n-1)d\]
Here:

\[a_n = nth \; term\]
a is first term of sequence,

n is number of terms,

and d is common difference..

Answer 47

As here you want to find 12th term so here n = 12

a = 6 and d = 10

Can you find \(a_{12}\).. ??

Answer 48

Oh sorry a2 is 6 and not a1..

Answer 49

Firstly find a1 or a :

\[a = a_2 - d\]

Answer 50

@shanesullins are you getting or not??

Please do interact otherwise I am unable to help you..

Answer 51

i got 6+(12-1)10 so far

Answer 52

No here second term is 6 not first term..

You have to find first term here firstly..

First term you can find by:

\[a = a_2 - d\]
Can you find a here??

Answer 53

-16

Answer 54

Yes now use the formula..

Answer 55

i got the wrong answer

Answer 56

Let me try..

Answer 57

\[a = a_2 - d \implies 6 - 10 = -4\]

Answer 58

\[a_{12} = -4 + (12-1)(10) \implies -4 + 110 \implies \color{blue}{a_{12} = 106}\]

Are not you getting this ??