Question

arithmetic sequcences and series

the fourth term is an arithmetic sequcence -6, and the 10th term is 5.
find the common difference and the first term.


how do i go about doing this? i tried so many numbers and the closest thing i got was 1/2 for the common difference :/

Answer 1

try this formula for an arithmetic sequence: \[\large a_n-a_k=d(n-k)\] use n=10, k=4, a10=5 and a4=-6 to solve the common difference

Answer 2

terms are
\[-6,-6+d,-6+2d,-6+3d,-6+4d,-6+5d,-6+6d=5\] solve for \(d\)

Answer 3

after solving d, use the same formula with n=10, a10=5, k=1 and the value of d to solve a1.

Answer 4

okay @sirm3d i got the a10=5 i get 2 and a4=-6 i got about -1 1/2

Answer 5

hmm i get \[-6+6d=5\]
\[6d=11\]
\[d=\frac{11}{6}\]

Answer 6

wait, what is the formula i should be using?
i have this one but i dont know how to use it:

avn = av1 + (n - 1)d

Answer 7

you can also use that formula. take n=10 and av10=5 and you'll have two unknowns av1 and d
use k=4 and av4=-6 and you'll get another equation in av1 and d.

Answer 8

\[\large n=10: av_{10} =av_1+(10-1)d\]\[\large n=4: av_{4}=av_1+(4-1)d\]

Answer 9

okay i get...

5= a + 9d and -6= a+3d

Answer 10

@sirm3d how can i figure out the rest?

Answer 11

multiply the second equation by (-1) then add it to the first equation. that should eliminate a and leave d to be solved.

Answer 12

\[\large \begin{matrix}5=a+9d \\ 6=-a-3d\end{matrix}\]

Answer 13

okay i get d=11/6

Answer 14

@sirm3d

Answer 15

is there anything else to do? i think there is but i dont know what

Answer 16

@Outkast3r09 is there anything i gotta do next ? and how?

Answer 17

@timo86m

Answer 18

now use d = 11/6 in either equation 1 or 2, then solve a
\[\huge 5=a+9(\frac{11}{6})\]

Answer 19

i get -1 1/2

Answer 20

\[\large 5 - \frac{33}{2} = a\]