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!

  • $\begingroup$ Just plug in $n=5$? $\endgroup$ – Don Thousand Apr 18 at 11:46
  • $\begingroup$ Sorry I made a mistake $\endgroup$ – T. Joel Apr 18 at 11:47


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 $7N+2:N+3$

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 $(7a + \frac{21}{2}r) = (30b + 45s)$

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 $65:12$.

  • $\begingroup$ Sorry the book's answer was 65:12? $\endgroup$ – T. Joel Apr 18 at 12:26
  • $\begingroup$ @T.Joel Made a mistake writing the general form of a sequence. It gets thrown off by whether you count the first term of a sequence as when $n=1$ or if you count the first term of the sequence as when $n=0$. $\endgroup$ – JMoravitz Apr 18 at 12:27
  • $\begingroup$ Is my own way of solving correct? $\endgroup$ – T. Joel Apr 18 at 12:39
  • $\begingroup$ @T.Joel yes, it looks good and frankly much easier and more straightforward than my own approach. Honestly, I had not seen the shortcut you used before for partial sums of arbitrary arithmetic sequences, only for the most basic of arithmetic sequences $1,2,3,\dots,n$. Knowing that the sum of $n$ consecutive terms in a sequence is the first term plus the last term times the number of terms divided by two is a very nice shortcut. $\endgroup$ – JMoravitz Apr 18 at 13:08
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    $\begingroup$ I still think my approach is a useful one to learn as it may be more applicable to other less specialized problems, but in terms of speed and efficiency your answer is better in my opinion. My thought process was to try to fill in whatever unknown information I could that might be useful and use that to reach a final conclusion. $\endgroup$ – JMoravitz Apr 18 at 13:11

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 $$\frac{1}{9}(7(9)+2):\frac{1}{9}(9+3) =65:12$$


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.


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