anonymous
  • anonymous
Find an equation for the nth term of the arithmetic sequence. a19 = -58, a21 = -164
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@Nnesha
Nnesha
  • Nnesha
hey:=) we need common difference and first term to write an equation arithmetic equation \[\huge\rm a_n = a_1 + (n-1)d\] where a_1 = first term d=common difference n= term that we have to find
Nnesha
  • Nnesha
given terms are a_{19}= -58 a_{21} = -164 plug this into the equation \[\large\rm a_\color{ReD}{{19}}=a_1 +(\color{ReD}{19}-1)d\] 19th term so we substitute n for 19 and a_{19} equal to -58 we can replace a_{19} with -58 \[\large\rm -58=a_1 +(\color{ReD}{19}-1)d\] same for a_{21} write an equation when n =21

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Nnesha
  • Nnesha
your turn:=) write an equation when n =21 just like i did for n=19 :=)
Nnesha
  • Nnesha
make sense ??
anonymous
  • anonymous
lets see, -164 = a1 (21 - 1)d ?
Nnesha
  • Nnesha
perfect! \[\huge\rm -164 = a_1 +(\color{Red}{21-1})d\] solve parentheses \[\large\rm -164 = a_1 +(\color{Red}{20})d\] this is our 2nd equation first one is \[\huge\rm -58 = a_1 +(\color{Red}{19-1})d\] \[\large\rm -58 = a_1 +(\color{Red}{18})d\]
Nnesha
  • Nnesha
\[\large\rm -58 = a_1 + \color{Red}{18} d\]\[\large\rm -164 = a_1 + \color{Red}{20} d\] we will use elimination method to solve this equation :=) so subtract equation one from equation 2 |dw:1444396691116:dw|
Nnesha
  • Nnesha
for elimination method we should change the sign of one of the equation in other words multiply by negative one now we have a_1 - a_1 we can cancel it out |dw:1444396859763:dw| now combine other terms
anonymous
  • anonymous
2d = -222 ?
Nnesha
  • Nnesha
hmm how did you get 2 d ?
Nnesha
  • Nnesha
remember we multiplied 2nd equation by negative so it's |dw:1444396988124:dw|
anonymous
  • anonymous
oops, i thought it was 18 okay so, 12d = -106 ?
Nnesha
  • Nnesha
right now solve for d
anonymous
  • anonymous
-8.833
Nnesha
  • Nnesha
hmm right but keep it in fraction |dw:1444397180870:dw| -53/6 now substitute this into the equation for d to find first term
Nnesha
  • Nnesha
\[\huge\rm a_n=a_1 +(n-1)d\] find first term substitute -53/6 for d
anonymous
  • anonymous
a_n = a_1 + (n - 1)-53/6 ... im kinda lost
Nnesha
  • Nnesha
hmm why ?
Nnesha
  • Nnesha
well n= number of term that we have to find so to find first term substitute n for 1
anonymous
  • anonymous
a_1 = a_1+ ( n - 1) -53/6
Nnesha
  • Nnesha
i seeeee
Nnesha
  • Nnesha
\[\large\rm -58 = a_1 + \color{Red}{18} d\]\[\large\rm -164 = a_1 + \color{Red}{20} d\] we can use one of these equation to find a_1 not original equation
anonymous
  • anonymous
Mathway.com is like a caculetor but is a web
Nnesha
  • Nnesha
lets use first one \[\rm a_1 +18d =-58\] substitute d for -53/6 to find a_1
Nnesha
  • Nnesha
i'm pretty sure i'm doing something wron ugh sorry let me see
anonymous
  • anonymous
i see, a_1 + 18(-53/6) = -58 would a_1 be -159 ..?
Nnesha
  • Nnesha
|dw:1444397762260:dw| you were right it's 18
anonymous
  • anonymous
ooooooooooooh lol!
Nnesha
  • Nnesha
i'm sorry. it was a_1 + (18 -1)d = a_1 + `18d`
Nnesha
  • Nnesha
now \[\rm a_1 +18(-53) = -58\]
anonymous
  • anonymous
okay sooo, a_1 + 18(-53) = -58 a_1 + -954 = - 58
anonymous
  • anonymous
a_1 = 896
Nnesha
  • Nnesha
looks good solve for a_1
anonymous
  • anonymous
soooo then would it be.. a_n = 896 -53 (n-1) ?
Nnesha
  • Nnesha
right :=)
anonymous
  • anonymous
woohooo! thank you so much!
Nnesha
  • Nnesha
np :=) btw you can distribute (n-1 ) by -53
Nnesha
  • Nnesha
\[\rm a_n = 896 -53n+53\] combine like terms

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