But it's too little information

Algebra Level 4

In a non-constant arithmetic progression of odd number of terms, having a positive integral common difference, the ratio of the sum of the first, third, fifth, seventh, ... terms to the sum of the remaining terms is 13:12. Then, what is the number of terms in this arithmetic progression?

The answer is 25.

1 solution

Ashish Menon
May 26, 2016

Let the first term of this AP be $a$ , the number of terms in this AP be $n$ and the common difference be $d$ .
Since the number of terms in this AP is odd, the first and the last terms are odd numbered terms. Now, the last term is $a + (n - 1)d$ .

Similarly, the second and the second last terms of this AP are even numbered term. So, there are $\dfrac{n + 1}{2}$ odd terms in this series. Similarly, there are $n - \dfrac{n + 1}{2} = \dfrac{n - 1}{2}$ .

Sum of the first, third, fifth, seventh, .... terms = $\dfrac{n + 1}{2 × 2} × \left(a + a+(n-1)d\right) = \dfrac{n + 1}{4} × \left(2a + (n-1)d\right)$ .
Sum of the second, fourth, sixth, eight, .... terms = $\dfrac{n - 1}{2 × 2} × \left(a + d+ a+(n-2)d\right) = \dfrac{n - 1}{4} × \left(2a + (n-1)d\right)$ .

Their ratio is equal to 13:12.

$\dfrac{\left(n + 1\right)\left(2a + (n - 1)d\right) × 4}{4 × \left(n - 1\right)\left(2a + (n - 1)d\right)} = \dfrac{13}{12}\\ \dfrac{n + 1}{n - 1} = \dfrac{13}{12}\\ 12\left(n + 1\right) = 13\left(n - 1\right)\\ 12n + 12 = 13n - 13\\ 13n - 12n = 12 + 13\\ n = \color{#69047E}{\boxed{25}}$

Nice solution. I solved it by assuming the number of terms be $2n-1$ . The solution was not as good. Ashish, I edited your problem. It should be "little" instead of "less". "Less" is for comparison. Also, don't leave a space between comma (,) and the word before it.

Chew-Seong Cheong - 5 years ago

Haha, I admit that gramatically I need to get a bit stronger. Thanks!

Ashish Menon - 5 years ago

Just upvoted your solution.

Chew-Seong Cheong - 5 years ago

@Chew-Seong Cheong – Thanks! The twist which I think in this question is that common difference remains d if we take the formula sum = $\dfrac{\text{Number of terms}}{2} × \left(\text{First term} + \text{Last term}\right)$ . Maybe the formula $\dfrac{\text{Number of terms}}{2} × \left(2 × \text{First term} + \left(\text{Number of terms} - 1\right) × \text{Common difference}\right)$ is a bit confusing in this case.

Ashish Menon - 5 years ago

If you assumed number of terms to be $2n-1$ , then $n ≥ 2$ , in addition to being a positive integer.

However, if you would have chosed number of terms to be $2n+1$ , then $n$ should be simply a positive integer.

Aditya Sky - 5 years ago

A much cleaner solution is to pair up the opposing ends. You can then conclude that the ratio of the number of terms is $13:12$ .

Calvin Lin Staff - 5 years ago

Yes, in my solution $\dfrac{n + 1}{n - 1} = \dfrac{13}{12}$ . Multiply and divide 2 on LHS. $\dfrac{\color{#3D99F6}{n + 1} × \color{#D61F06}{2}}{\color{#3D99F6}{2} × \color{#D61F06}{n - 1}} = \dfrac{13}{12}$ . But $\dfrac{n + 1}{2}$ is the number of odd numbered terms and $\dfrac{n - 1}{2}$ is even numbered terms as stated in my solution, so $\dfrac{\text{Number of odd numbered term}}{\text{Number of even number terms}} = \dfrac{13}{12}$ .

Ashish Menon - 5 years ago

But it's too little information

The answer is 25.

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