Let $\{a_n\}$ and $\{A_n\}$ be two arithmetic sequences, and $s_n$ and $S_n$ sums of ther first $n$ terms.
$$If\ \ \frac{s_n}{S_n}=\frac{7n+1}{4n+27},\ calculate\ \ \frac{a_{11}}{A_{11}}$$.
Let $\{a_n\}$ and $\{A_n\}$ be two arithmetic sequences, and $s_n$ and $S_n$ sums of ther first $n$ terms.
$$If\ \ \frac{s_n}{S_n}=\frac{7n+1}{4n+27},\ calculate\ \ \frac{a_{11}}{A_{11}}$$.
$$\frac{a_{11}}{A_{11}}=\frac{a_1+10d}{A_1+10D}$$
I supose problem must transform in system of 3 equation with 3 unknown, $a_1$, $d$, and $D$, ($d,D$ coresponding differences)
Hint:
nth term:
a + d(n-1)
A + D(n-1)
So 11th term:
a + 10d
A + 10D
Also, using sum of 1st nth terms:
(2a + dn - d)/(2A + Dn - D) = (7n + 1)/(4n + 27)
Now substitute n = 11
Here's my approach:
Since $\displaystyle \frac{s_n}{S_n} =\frac{7n+1}{4n+27}$
then let $\displaystyle s_n = k(7n+1)$ and $\displaystyle S_n = k(4n+27)$
Then use the fact that $\displaystyle a_1 = s_1$ and $\displaystyle a_2 =s_2 - s_1$ to find d.
Do the same for A, S and D.
Then see what you can do.
Do we have the whole question? I have far too many variables at the end and not enough conditions to specify them.
ala DenisB:
For the series $\displaystyle a_1,~a_1 + d,~a_1 + 2d,~ \text{...}$
we have $\displaystyle s_k = \dfrac{k}{2} (a_1 + a_k)$
So for k = 11: $\displaystyle s_{11} = \dfrac{11}{2} ( a_1 + a_{11})$
Similarly for the A series:
$\displaystyle S_{11} = \dfrac{11}{2} (A_1 + A_{11})$
and we have
$\displaystyle \dfrac{s_{11}}{S_{11}} = \dfrac{a_1 + a_{11}}{A_1 + A_{11}} = \dfrac{7 \cdot 11 + 1}{4 \cdot 11 + 27} = \dfrac{78}{71}$
Cross mulitplying, grouping, etc. gives
$\displaystyle \dfrac{a_{11}}{A_{11}} = \dfrac{78}{71} \left ( \dfrac{A_1}{A_1 + 10D} \right ) - \left ( \dfrac{a_1}{A_1 + 10D} \right ) + \dfrac{78}{71}$
Oddly enough this is independent of d. But still, we have three unknowns here: $\displaystyle a_1,~A_1,~D$. I can't go any further.
-Dan
Condition
$$ \frac{s_n}{S_n} =\frac{7n+1}{4n+27}$$
holds for every n. I think it must be used for n=1,2,3,4 to give enaf equations, but system will be homogenous, and this is a way I hate. I hope it exist some more elegant way. I don't have solution, only answer $\frac{a_{11}}{A_{11}}=\frac{4}{3}$.