Let ${a_n}$ and ${b_n}$ be two different arithmetic series such that $(3n+1)a_n=(2n-1)b_n$ for all positive integers $n$. Let $A_n=a_1+a_2+...+a_n$ and $B_n=b_1+b_2+...+b_n$. what is the value of $\frac{A_9}{B_6}$?
How do I do this? There isn't any info about the separate sequences, but how they are related. Do I need to know anything about the sequences themselves?
On
Another approach is to notice that all the elements of both series scale with $a_1$, so we can just set $a_1=1$. Let the difference of the $a$ series be $k_a$ and the difference of the $b$ series be $k_b$. Then we can just write the equations for $n=1,2,3$ and get $$4=b_1\\7a_2=3b_2\\10a_3=5b_3 \\7(1+k_a)=3(b_1+k_b)\\10(1+2k_a)=5(b_1+2k_b)\\k_a=2\\k_b=3\\a_n=1+2(n-1)\\b_n=4+3(n-1)$$ and sum the series to get $$A_9=81,B_6=69$$
On
Note for any two arithmetic sequences $a_i = a + i*d$ and $b_i = b + ie$ you have $\frac {a_n}{b_n} = \frac{a+ nd}{b+ ne}= \frac{r(a+ nd)}{r(b+ne)}$
So we have $\frac {a_n}{b_n} = \frac {2n-1}{3n+1}=\frac {r(a + nd)}{r(b+ne)}$
So $a_i = -k +i*(2k)$ and $b_i = k + i*(3k)$ for some real value $k=1/r$.
So $A_n = \sum a_i = -nk + 2k\frac{n(n+1)}2$ and $B_n = \sum b_i = nk + 3k\frac{n(n+1)}2$
and $\frac {A_n}{B_n} = \frac {-nk + kn(n+1)}{nk + 3kn\frac {n+1}2}=\frac{n}{3n + 2.5}$
$a_n$ and $b_n$ are both arithmetic so can be wrtten in the form ... $$ a_n=k_a+d_an \\b_n=k_b+d_bn $$
$$ (3n+1)a_n = (2n-1)b_n \\ \implies 3d_a n^2 + (3K-a + d_a)n +k_a = 2 d_b n^2+(2k_b-d_b)n -k_b $$
This is an identity in $n$ , so we can equate coefficients of powers of $n$ to get a system of 3 equations and 4 unknowns ...
$$\begin{eqnarray*} k_a &=& -k_b \\3k_a+d_a &=& 2 k_b-d_b \\3d_a &=& 2 d_b \end{eqnarray*}$$
which simplifies to ... $$\begin{eqnarray*} k_a &=& -k_b \\d_b &=& 3 k_b \\d_a &=& 2 k_b \end{eqnarray*}$$
Now let $x \equiv k_b$ so that the sequences can be expressed as ... $$a_n=x(-1+2n) \\ b_n = x(1+3n)$$
Hopefully you can take it from here, $x$ will cancel when you take the ratio of the sums.