Finding the ratio of 5th of two different arithmetic sequences

Question

It is given that the ratio of the sum to the nth term of two different arithmetic sequences is $7n+2:n+3$. Find the ratio of the 5th term of the sequences. I have no idea where to start this pls help!

Answer 1

Setup:

Let the first sequence be $a+r(n-1)$ and let the second sequence be $b+s(n-1)$.

The partial sum of the first $N$ terms of these will be $aN + r\frac{N(N-1)}{2}$ and $bN + s\frac{N(N-1)}{2}$ respectively.

By factoring out an $N$ from both expressions, we find that the ratio of partial sums is $a+r\frac{N-1}{2}:b+s\frac{N-1}{2}$

We are told from the problem statement that this ratio can also be written as $7N+2:N+3$

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For two ratios $A:B$ and $C:D$ to be equivalent, this means by definition that $AD=BC$ so by applying this to our problem, we learn that:

$$(a+r\frac{N-1}{2})(N+3) = (b+s\frac{N-1}{2})(7N+2)$$

This should hold for all values of $N$, including for example $N=1$. From $N=1$ we learn that $(4a) = (9b)$.

From using $N=2$ we learn that $(5a+\frac{5}{2}r) = (16b + 8s)$

From using $N=3$ we learn that $(6a + 6r) = (23b + 23s)$

Finally, from using $N=4$ we learn that $(7a + \frac{21}{2}r) = (30b + 45s)$

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Now, with all of this we have found four equations involving our four unknowns $a,b,r,s$.

If there was a unique solution for $a,b,r,s$ we would be able to find them using Gaussian Elimination or Row Reduction, etc...

As it so happens, we cannot determine them uniquely, however we can determine them based on a single parameter which is good enough since we are not interested in the exact values of the sequences but rather the ratio of the fifth terms of the sequences.

We learn from row reduction that by letting $s$ be equal to a parameter $t$ that we have $(a,r,b,s) = (9t/2, 7t, 2t, t)$ and so the fifth terms of the sequences would be $9t/2 + 28t$ and $2t + 4t$ respectively and the ratio will be $65:12$.

Answer 2

The 5th term, $a_5$ of any arithmetic sequence is the arithmetic mean of the 1st term, $a_1$ and the 9th term, $a_9$. So we get $2a_5=a_1+a_9$ The sum to the 9th term is also $\frac{9}{2}(a+a_9)=9a_5$ So we get $$S_9=9a_5$$ $$a_5=\frac{1}{9}S_9$$ So the ratio of the sums to the 9th term is $$\frac{1}{9}(7(9)+2):\frac{1}{9}(9+3) =65:12$$

Answer 3

Starting from JMoravitz's answer, consider $$(a+r\frac{N-1}{2})(N+3) - (b+s\frac{N-1}{2})(7N+2)=0$$ Expand and group terms to get $$\left(3 a-2 b-\frac{3 }{2}r+s\right)+ \left(a-7 b+r+\frac{5 }{2}s\right)N+\frac{1}{2} (r-7 s)N^2=0$$ and each coefficient must be $0$; so, three linear equations in $(a,b,r,s)$.

As also said in the same answer, using $N=1$ leads to $4a-9b=0$ and then the solution.