Pre-‐Calculus Name: ANSWER KEY Unit 1 Arithmetic, Geometric Sequences And Series REVIEW Period: _________ Date: _________________ Arithmetic Series Geometric Series Infinite Series
!!!!
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an = a1 + (n−1)dSn = n
2 a1 + an( )
an = a1rn−1
Sn =a1(1− r
n )1− r
!!!!
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S =a1
1− r
1) For the sequence 10,$6,$2,$−2,$... find each of the following: Note this is an arithmetic sequence with d = -‐4.
a) Recursive formula: 1a) a1=10; an+1=an -‐4 an+1=an + d; so an+1 = an – 4. And a1 = 10.
Explicit formula: 1b) an= -‐4n+14 b)
an=a1+d(n-‐1). So, an=10+(-‐4)(n-‐1) an=10-‐4n+4 = -‐4n+14
c) The 15th term 1c) a15=-‐46 a15= -‐4(15) +14 (this is from part b) so, a15=-‐46
d) The sum of the first 15 terms 1d) -‐270 𝑆! =
!!𝑛(𝑎! + 𝑎!)
We know a1=10, and a15=-‐46 from part c). So, 𝑆!" =
!!15 10− 46 = −270.
2) Find d for the arithmetic sequence in which !!!!
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a1 = 6 and !!!!
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a17 =−42. 2) -‐3 Use an=a1+d(n-‐1). We know a1=6 and a17=-‐42 and n=17. So, -‐42 = 6 + d(17-‐1). Then -‐42 = 6 + d(16). (subtract 6 from both sides) -‐48 = d(16). (divide both sides by 16) And finally -‐3 = d.
3) For the geometric sequence !!
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127, ! 1
9, ! 1
3, ! ...find each of the following:
Se see that r = 3, and that a1= !!"
a) Recursive formula: 3a) a1= !!", an+1 = an(3)
b) Explicit formula: 3b) 𝑎! =!!"(3!!!)
𝑎! = 𝑎!(𝑟!!!)
c) The 14th term 3c) 59049 𝑎!" =
!!"
3!" = 59049 d) The sum of the first 10 terms 3d) 1093.48
𝑆! =!! !!!!
!!!. 𝑆𝑜 𝑆! =
!!" !!!
!"
!!! = !
!"(29524) = 1093.48
4) If r = 2 and a4 = 28, what is the first term of the sequence? 4) a1= 3.5 𝑎! = 𝑎! 𝑟!!! , 𝑎𝑛𝑑 𝑎! = 28. 28 = 𝑎! 2! . 𝑆𝑜,𝑎! =
!"!= 3.5
You can also back-‐track from a4 to a1 by dividing by r three times. a4=28, so a3=14, then a2=7, and finally a1=3.5. 5) Find the sum of the series (if it exists) !!
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18+ 6+ 2+ ... 5) Sum = 27 This is a geometric series with r = !
! and a1=18.
Since |r| < 1. We can find the sum of the infinite series (it converges) So 𝑆! =
!!!!!
. 𝑓𝑖𝑙𝑙 𝑖𝑛 𝑘𝑛𝑜𝑤𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 → 𝑆! =!"
(!!!!)= !"
!!= 27.
6) Do the following series appear to converge or diverge? a) 0.2+0.02+0.002+... 6a) Converges r = 0.1 b) 2
7+ 4
7+ 8
7+... 6b) Diverges
r = 2
7) Find the value of 2n( )n=1
5
∑ . 7) 30
Since there are only 5 values (n=1 to 5). We can find each term separately then add up. n=1 gives 2(1) = 2 n=2 fives 2(2) = 4 n=3 gives 2(3) = 6 n=4 gives 2(4) = 8 n=5 gives 2(5) = 10 Add up 2+4+6+8+10 = 30. Can also use partial sums formula for arithmetic sequence (d=2). a1 = 2, a5 = 10 So, 𝑠! =
!!(5) 2+ 10 = 30.
8) Express the series 3+ 9+ 27+81+... using sigma notation. 8) See Below This is a geometric series with a1= 3, and r = 3. It is infinite since it has … at the end. Since it is infinite it will go from k = 1 to infinity. (make sure variable matches) Left part of the Sigma Notation will be
_____!
!!!
We put the explicit form of the geometric sequence to the right of the sigma. Explicit form: 𝑎! = 𝑎! 𝑟!!! . 𝑊𝑒 𝑘𝑛𝑜𝑤 𝑎! 𝑖𝑠 3 𝑎𝑛𝑑 𝑟 𝑖𝑠 3. 𝐸𝑥𝑝𝑙𝑖𝑐𝑖𝑡 𝑓𝑜𝑟𝑚 𝑎! = 3 3!!! . Since 3 𝑥 3!!! have the same base of 3, we can simplify into 3n . So the overall expression will be
3(3!!!!
!!!
) 𝑂𝑅 3!!
!!!
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