Question

explicit/recursive problem: pls read further

Answer 1

ok,
\(\large a_n=13+(n-1)6\)
can you find \(a_{n-1}\) from this ?
(hint: replace 'n' by 'n-1)

Answer 2

n-2??

Answer 3

just n-2 ?

replace 'n' by n-1 in the entire equation and tell me what u get ?

Answer 4

what do u mean replace "n"??

Answer 5

like if an was
an = 4n+9
then a(n-1) will be 4 (n-1) +9
like wherever u see "n" just write "n-1" instead

Answer 6

it is c.d i.e.,6

Answer 7

yeah, but there's a method to get there...

Answer 8

every term of an A.P increases by c.d
an=an-1+c.d

Answer 9

so then what?

Answer 10

so, your \(a_{n-1} = 13+(n-2)6\)
and you just do a simple subtraction of \(a_n-a_{n-1}\)
to get your answer

Answer 11

wait so the n-1 became n-2 ?

Answer 12

yes, because i replce n by n-1
(n-1) -1 = n-2

Answer 13

right ok & then solve

Answer 14

yes, just subtract, and notice the n terms getting cancelled

Answer 15

-1/6?

Answer 16

that's not right :(

Answer 17

how u got -1/6 ?

Answer 18

im sorry out of all the units this is the one im most confused w/: series & sequences & stuff

Answer 19

ok so its set up as this right:?
a n-1= 13+ (n-2)6

Answer 20

yes, now an - a n-1 = 13+ (n-1)6 - ( 13 + (n-2)6 ) =... ?

Answer 21

ok wait what did u get

Answer 22

i got 6, did u get the same ?

Answer 23

no... I got this weird ans.. this should be easy but I think im overthinking it :( :(

Answer 24

13+ (n-1)6 - ( 13 + (n-2)6 )
= 13 + 6n- 6 - (13+ 6n-12)
ok till here ?

Answer 25

yea

Answer 26

got that part

Answer 27

so,
=13 + 6n- 6 - 13- 6n+12
what gets cancelled ?

Answer 28

the 2nd part would be minus 13 PLUS 6n wouldn't it?

Answer 29

nopes, the - *minus* distributes to all 3 terms
- (13+ 6n-12) = -13 -6n +12

Answer 30

ohh right cuz its like a -1

Answer 31

yes :)

Answer 32

ok so yea... i guess i got 6 too

Answer 33

i'm gonna practice more of these when i get more time so i wont do so bad on a final test

Answer 34

yes, practice will make things easier :)

Answer 35

thanks so much hartnn

Answer 36

welcome ^_^

Answer 37

I get blank =\[ -a_{n}+1+13+(n-1)6\]

Answer 38

& then? what dod U get as ur final ans?

Answer 39

?

Answer 40

-18

Answer 41

really?

Answer 42

what ? how? it isn't -18 it is 6, the common difference,
even if u think logically, the difference between current term an and previous term a n-1 is always the common difference, here = 6

Answer 43

Because \[a _{n}-a _{n-1}=13+(n-1)6-(13+(n-1 -1))6\]
So \[a _{n}= a _{n-1}+ (n-1)6 - (n-2)6\]
Therefore,\[6(n-1 - (n-2))\], which is actually 6, not -18.

Answer 44

so u say 6 too? :p

Answer 45

yep

Answer 46

ok :)

Answer 47

defintely

Answer 48

definitely*

Answer 49

triple confirmation, lol ....4th confirmation coming...

Answer 50

:) I get it, it is 6

Answer 51

I disagree