what is the 7th term in the sequence a(1 )=12 a(n)=a(n)-1+4 I need help finding the recursive formula.
Accepted Solution
A:
Answer:Explicit form: [tex]a_n=8+4n[/tex]Seventh term: [tex]a_7=36[/tex]You gave the recursive form:[tex]a_n=a_{n-1}+4 \text{ with } a_1=12[/tex]Step-by-step explanation:You already have the recursive formula which is:[tex]a_n=a_{n-1}+4[/tex] where [tex]a_1=12[/tex].Maybe you looking for the explicit form and also the 7th term?You were just looking for the 7th term. Sometimes you can read the recursive formula pretty easily and understand the pattern that is happening.[tex]a_{n-1}[/tex] is the term right before [tex]a(n)[/tex]. Just like [tex]a_5[/tex] would be the term right before [tex]a_6[/tex].Anyways becak to our recursive for a sequence that was given:[tex]a_{n}=a_{n-1}+4[/tex] says term=previous term+4.So you adding 4 over and over to generate the terms of a sequence. This is arithmetic sequence because it is going up by same number (or could go down by same number). The common difference is 4.[tex]a_1=12[/tex][tex]a_2=12+4=16[/tex][tex]a_3=16+4=20[/tex][tex]a_4=20+4=24[/tex][tex]a_5=24+4=28[/tex][tex]a_6=28+4=32[/tex][tex]a_7=32+4=36[/tex]Now if you don't like that. You could just blindly without trying to understand the meaning of it just plug numbers in:[tex]a_1=12[/tex]For if we wanted to know the 2nd term; we would plug in 2 for n:[tex]a_2=a_{2-1}+4[/tex][tex]a_2=a_1+4[/tex][tex]a_2=12+4[/tex][tex]a_2=16[/tex]Plug in 3 for the 3rd term:[tex]a_3=a_{3-1}+4[/tex][tex]a_3=a_2+4[/tex][tex]a_3=16+4[/tex][tex]a_3=20[/tex]Plug in 4 for the 4th term:[tex]a_4=a_{4-1}+4[/tex][tex]a_4=a_3+4[/tex][tex]a_4=20+4[/tex][tex]a_4=24[/tex]Plug in 5 for the 5th term:[tex]a_5=a_{5-1}+4[/tex][tex]a_5=a_4+4[/tex][tex]a_5=24+4[/tex][tex]a_5=28[/tex]Plug in 6 for the 6th term:[tex]a_6=a_{6-1}+4[/tex][tex]a_6=a_5+4[/tex][tex]a_6=28+4[/tex][tex]a_6=32[/tex]Plug in 7 for the 7th term:[tex]a_7=a_{6-1}+4[/tex][tex]a_7=a_5+4[/tex][tex]a_7=32+4[/tex][tex]a_7=36[/tex]Now if you look at the points we just got (I will just go up to 4 terms):n (treat as x) | 1 2 3 4 a(n) (treat as y| 12 16 20 24The explicit form in not in terms of other terms of the sequence. You are looking here for an equation that relates n (x) to a(n) (y).I'm just going to use x and y until the end where I will replace them back in terms of n and a(n).This is a line because the's rise/run (the slope) is the same per choosing of points.That is the following is true:[tex]\frac{16-12}{2-1}=\frac{20-16}{3-2}=\frac{24-20}{4-3}[/tex] son on....These all the the value 4, the same number that the arithmetic sequence is going up by. So in an arithmetic sequence, the common difference is the slope of the line.I'm going to use point-slope formula which is [tex]y-y_1=m(x-x_1)[/tex] where [tex](x_1,y_1)[/tex] is a point on the line and [tex]m[/tex] is the slope.So we have m=4 and [tex](x_1,y_1)[/tex] could be any of the 7 we found, but I will choose (1,12) for it:[tex]y-12=4(x-1)[/tex]Add 12 on both sides:[tex]y=12+4(x-1)[/tex][tex]a_n=12+4(n-1)[/tex].This is actually in the form of most books formulas for an explicit form which is [tex]a_n=a_1+d(n-1)[/tex] where [tex]a_1[/tex] is the first term and d is the common difference. So another way to do the problem:You would have to know the following are equivalent:[tex]a_n=a_{n-1}+d[/tex] with [tex]a_1 \text{ is given}[/tex]and[tex]a_n=a_1+d(n-1)[/tex].If you know these are equivalent then you could compare [tex]a_n=a_{n-1}+d[/tex] to [tex]a_n=a_{n-1}+4[/tex] and determine d is 4.You could also see that [tex]a_1[/tex] is give as 12.Then you just plug into:[tex]a_n=a_1+d(n-1)[/tex][tex]a_n=12+4(n-1)[/tex].You could also simplify this equation just a bit.You could distribute and then combine a pair of like terms.Like so,[tex]a_n=12+4(n-1)[/tex][tex]a_n=12+4n-4[/tex][tex]a_n=8+4n[/tex]