Question

Question.

Answer 1

Answer.

Answer 2

Here's the real question:

If the ratio of the sum of two arithmetic progressions up to the \(n\)th terms is \(\dfrac{5n + 4}{9n + 6}\), find the ratio of the \(13\)th terms. Also derive the expression for the ratio of \(n\)th terms.

Answer 3

I know the solution to this, but I was wondering if we could also use \(S_{n} - S_{n - 1} = a_n\).

Answer 4

@BSwan

Answer 5

I know the solution to this. Just wanted to know if there was an alternative.

Answer 6

\[A=\{a_0,a_1,a_2,...,a_n\}\]\[B=\{b_0,b_1,b_2,...,b_n\}\]
\[S_n=a_0+b_0+a_1+b_1+a_2+b_2+...+a_n+b_n\]
im assuming this is what its refering to?

Answer 7

\[S_n=(a_0)+(a_0+d)+(a_0+2d)+...+(a_0+dn)\\
~~~~~+(b_0)+(b_0+k)+(b_0+2k)+...+(b_0+kn)\]

\[S_n=a_0(n+1)+(d+2d+...+dn)\\
~~~~+b_0(k+1)+(k+2k+...+kn)\]

\[S_n=a_0(n+1)+d\frac{n(n+1)}{2}+b_0(k+1)+k\frac{k(k+1)}{2}\]

Answer 8

nonono

Answer 9

little mix up in the coding ... bo(n+1) + kn(n+1)/2 ....

\[S_{n+1}=[2a_0(n+2)+dn(n+2)+2b_0(n+2)+kn(n+2)]/{2}\]\[S_n=[2a_0(n+1)+dn(n+1)+2b_0(n+1)+kn(n+1)]/{2}\]


\[S_{n+1}=[2a_0(n+2)+dn(n+2)+2b_0(n+2)+kn(n+2)]\]\[/S_n~~=[2a_0(n+1)+dn(n+1)+2b_0(n+1)+kn(n+1)]\]

Answer 10

There are two different arithmetic progressions \(t_n \) and \(T_n\).\[\large \frac{\sum t_n}{\sum T_n} = \dfrac{5n +4}{9n + 6}\]This relation is true for all \(n\) where \(n\) is the \(n\)th term.

Answer 11

\(\large \dfrac{S_n}{S_n^{'}} = \dfrac{5n+4}{9n+6}\)


\(\large \dfrac{a_n}{a_n^{'}} = \dfrac{S_n - S_{n-1}}{S_n^{'} - S_{n-1}^{'}} = \cdots = \dfrac{10n - 1}{18n-3}\)


plugin n = 13, you get ratio of 13th terms = 43/77

Answer 12

but i have another method(u must be having this i think...) which is simpler than this...

Answer 13

so: sumA/sumB as opposed to what i had read it to be

Answer 14

How did you get that expression? Can you fill some more steps?

Answer 15

\(\dfrac{a_n}{a_n^{'}} = \dfrac{S_n - S_{n-1}}{S_n^{'} - S_{n-1}^{'}} = \dfrac{n(5n+4) - [(n-1)(5(n-1)+4)]}{n(9n+4) - [(n-1)(9(n-1)+4)]}=\cdots = \dfrac{10n - 1}{18n-3}\)

Answer 16

http://www.wolframalpha.com/input/?i=simplify+%28n%285n+%2B+4%29+-+%28n-1%29%5B5%28n-1%29+%2B+4%5D%29%2F%28n%289n+%2B+6%29+-+%28n-1%29%5B9%28n-1%29%2B6%5D%29

Answer 17

I was hesitating to plug in \(S_n\) and \(S_n\) because I thought those were just the ratios, but it makes sense to me now. lol

What's your easier method?

Answer 18

key thing is to see that "n" cancels out in the ratio of sums, but it wont cancel out when u take the ratio of difference of two successive sums, other common factors might cancel out... so the numerator and denominator terms dont exactly give us the series

Answer 19

Yes, I see that. :)

What's the easy method though?

Answer 20

Since the left side is also a ratio, there is nothing in writing S_n/S_n'

Answer 21

I believe my other method same as the solution known to you... it converts the ratio of terms to a sum ratio and compares the coefficients

Answer 22

Still... how does it do that?

Answer 23

ratio of 13th terms :

a1 + 12d1
-------
a2 + 12d2

Answer 24

given ratio of sums :

2a1 + (n-1)d1 5n+4
------------- = ----------
2a2 + (n-1)d2 9n+6

Answer 25

haha yup, that's the one

Answer 26

ikr... :)

Answer 27

thank you very much!

i'm pretty sleepy now

Answer 28

np, have good sleep :)

Answer 29

btw, thanks for the starting idea \(\large S_n - S_{n-1 } = a_n\) :)

Answer 30

Zenos xD