Question

an arithmetic sequence has a1 = a5 and a5 = 13. how many terms of this sequence must be added to obtain 60?

Answer 1

if a_1=a_5 and a_5=13
then a_1=13
so sounds like your common difference is 0

Answer 2

I'm afraid you will never be able to get exactly 60 by adding a some of the terms of the sequence

Answer 3

mainly because 13 doesn't divide 60

Answer 4

unless there is a mistake in the question

Answer 5

maybe it wants us to get to the number n such that the sum is at least 60
a1+a2+a3+...+an=60
and that case we could do it
so you have 13+13+13+...+13=13n
solve 13n=60 for n

Answer 6

I meant a_1=5 and a_5=13

Answer 7

you can find the common difference by doing
\[\frac{a_5-a_1}{5-1}\]

Answer 8

let me know when you think you have the common difference

Answer 9

2?

Answer 10

right

Answer 11

\[\text{ the sequence looks like } \\ 5,7,9,11,13,15,17,19,...\]

Answer 12

5+7+9+11+13+15=?

Answer 13

there is another way to do this
if you don't like that

Answer 14

so a_6?

Answer 15

well it asked for how many terms we need to add

Answer 16

so close
you needed 6 of the first terms

Answer 17

so 6 terms?

Answer 18

just adding the terms til you get to 60

Answer 19

\[Sum=\frac{n (a_1+a_n)}{2} \\ Sum=\frac{n(a_1+(a_1+(n-1)d))}{2} \\ \text{ you said } d=2 \text{ and } a_1=5 \text{ and we want sum=60 }\]
\[60=\frac{n(5+(5+(n-1)2))}{2} \\ \text{ \times 2 on both sides } \\ 120=n(5+5+(n-1)2) \\ 120=n(10+2n-2) \\ 120=n(8+2n) \\ 120=2n^2+8n \\ \\ 2n^2+8n-120=0 \\ n^2+4n-60=0 \\ (n-6)(n+10)=0\]

Answer 20

but you also get n=6 from this way
n=-10 by the way doesn't make sense

Answer 21

Anyways the way that uses a quadratic equation might be better for bigger numbers

Answer 22

I will make it look a little pretty
the formula that is that I used
\[Sum=\frac{n(2a_1+(n-1)d)}{2} \\ Sum=\frac{n(2a_1+dn-d)}{2} \\ Sum=\frac{2 a_1 n +dn^2-dn}{2} \\ 2 Sum=2a_1 n+dn^2-dn \\ 0=2a_1n+dn^2-dn-2 Sum \\ 0=dn^2+(-d+2a_1 )n -2 Sum\]

Answer 23

so you can use that bottom quadratic to find n

Answer 24

should be factorable if there is an integer such that allows you to add a certain amount of terms together to get a certain sum

Answer 25

\[n=\frac{d-2a_1 \pm \sqrt{(-d+2a_1)^2-4(d)(-2Sum)}}{2d} \\ n=\frac{d-2a_1 \pm \sqrt{(-d+2a_1)^2+8d Sum}}{2d}\]
but if you don't like factoring
you could use the quadratic formula as I did here
one of the answers will not work just like I have shown about with the n=-10 which did not make sense :)

Answer 26

ok thank you i think i got it now

Answer 27

but hey I know that probably looks a little confusing
pretend we have the arithemetic sequence
7,10,13,16,...
common difference d=3
a_1=7
pretend we want sum 30
plug in where have that formula for n

Answer 28

whoa nelly is this the problem \(a_1=a_5\)?

Answer 29

of it is arithmetic, and \(a_1=a5\) then all the terms must be the same right?

Answer 30

\[n=\frac{3-2(7) \pm \sqrt{(-3+2(7))^2+8(3)(30)}}{2(3)} \\ n=\frac{-11 \pm \sqrt{(11)^2+720}}{6} \\ \text{ going \to throw out the obvious negative value } \\ n=\frac{-11 + \sqrt{121+720}}{6} \\ n=\frac{-11 + \sqrt{841}}{6}\]
\[n=\frac{-11+29}{6}=\frac{18}{6}=3\]

Answer 31

which is true since 17+10+13=30

Answer 32

anyways yeah he said that was a typeo earlier

Answer 33

oh i came late

Answer 34

\[5+7+9+11+13+...=60\]

Answer 35

i prefer that way of adding up the terms til you get to 60

Answer 36

\[\sum_{k=1}^n2k-1=n^2\]

Answer 37

but if the sum you want is large you probably have to do it the icky way

Answer 38

you have \(60\) and are missing \(1\) and \(3\) so you have 8 terms

Answer 39

oh you want to do it that way
thats cute too

Answer 40

you are cute as a unicorn

Answer 41

i wouldn't know anything about unicorns

Answer 42

https://www.youtube.com/watch?v=oDFiri7HH1E