Just arithmetic

Algebra Level 3

An arithmetic progression consisting of $n$ terms which are alternatively odd and even has its first term $a$ an odd positive integer and common difference $d$ also a positive integer.

When the sum of the first, third, fifth, seventh, ... terms are subtracted from the sum of the second, fourth, sixth, eight, ... terms, the result obtained is 501. Find $n × d$ .

The answer is 1002.

2 solutions

Abhay Tiwari
Jun 8, 2016

The difference between two consecutive terms in an AP is $d$ , which we call the common difference.

Now, total number of terms are $n$

$\underbrace{(2^{nd} \space term + 4^{th} \space term + 6^{th} \space term + \dots + n^{th} \space term)}_{\frac{n}{2} \space terms}-\underbrace{(1^{st} \space term + 3^{rd} \space term + 5^{th} \space term + \dots + (n-1)^{th} \space term)}_{\frac{n}{2} \space terms}=501$ .

$(2^{nd} \space term -1^{st} \space term )+(4^{th} \space term-3^{rd} \space term)+(6^{th} \space term-5^{th} \space term )+\dots+( n^{th} \space term-(n-1)^{th} \space terms)=501$

$\underbrace{d+d+d+\dots+d}_{\frac{n}{2} \space d's}=501$

$\frac{n}{2} × d=501$

$n × d=\color{#D61F06}{\boxed{1002}}$ !

Well, maybe $-$ should be substituted by $\dfrac{n}{2}$ , right?

Ashish Menon - 5 years ago

For which line are we talking about?

Abhay Tiwari - 5 years ago

Maybe this is my device's problem!

Ashish Menon - 5 years ago

@Ashish Menon – This is interesting.

Abhay Tiwari - 5 years ago

@Abhay Tiwari – Woah, a bug!!! Remove the image from your solutioh, it makes it look wierd XD

Ashish Menon - 5 years ago

@Ashish Menon – Sometimes this site behaves abnormally. It gets on my nerve. For instance, that pic I wanted to upload for replying you. And it is now stuck there like a vaccum pump.

Abhay Tiwari - 5 years ago

@Abhay Tiwari – Haha, yeah true. It happens that the page reloads a lot of time erasing away all the work I did even if it was a huge one and I have to do it all over again. -.- :P

Ashish Menon - 5 years ago

@Ashish Menon – Many a time I post a question, and my previous question gets posted with the answer of my new questions. And that is really weird. Make me feel like I am goofing around here XD

Abhay Tiwari - 5 years ago

@Abhay Tiwari – Yeah, true

Ashish Menon - 5 years ago

@Abhay Tiwari – Hmm.. nice solution(+1)

Ashish Menon - 5 years ago

Ashish Menon
May 27, 2016

This AP starts with an odd positive integer and its term are odd and even alternatively. So, we can comment that the common difference is odd. Now, the difference between the sum of the even numbered terms and the sum of the odd numbered terms is $501$ indicating that the sum of the even numbered terms is greater than the sum of the odd numbered terms. Since, these terms are in AP of alternating even and odd terms, we can comment that this AP ends with an even number. This is because if it was ending with an odd number then the sum of the odd numbered terms would have been greater than the sum of the even numbered terms. Since this AP ends with an even number, we can comment that the number of terms in this AP is also even because this AP of alternating odd and even numbers start with an odd positive integer.

So, the number of odd numbered terms = number of even numbered terms = $\dfrac{n}{2}$ .
Sum of odd numbered terms = $\dfrac{n}{2 × 2} × \left(a + a + (n - 2)d\right) = \dfrac{n}{4} × \left(2a + nd - 2d\right)$ .
Sum of even numbered terms = $\dfrac{n}{2 × 2} × \left(a + d + a + (n - 1)d\right) = \dfrac{n}{4} × \left(2a + nd\right)$ .
Sum of even numbered term - Sum of odd numbered terms = $501$ .
$\left(\dfrac{n}{4} × \left(2a + nd\right)\right) - \left(\dfrac{n}{4} × \left(2a + nd - 2d\right)\right) = 501\\ \dfrac{n}{4} × \left(2a + nd - 2a - nd + 2d\right) = 501\\ n × 2d = 2004\\ nd = \dfrac{2004}{2}\\ n × d = \color{#69047E}{\boxed{1002}}$

The common Difference will be odd not even..Please see to it.. And I don't understand one thing.. How is sum of even numbered terms = n(2a+nd)/4? I mean S = n/2 × [2a + (n-1)d].. And n will be equal to n/2.. So how will it be equal to n(2a+nd)/4? Same confusion goes for odd numbered sum..

Sagar Shah - 5 years ago

I saw into it, the common difference is odd and not even XD I edited it. Anyways its of no use. If we are considering only even numbered term, there are $\dfrac{n}{2}$ right. Now the first even number is $a_2 = a+d$ because it starts with an odd positive integer. Since this AP have alternating odd and even terms, the second term would be even. The last term is even as proved in the solution. The last term of the original sequence can be written as $a + (n - 1)d$ . Now, sum of even numbered terms = $\dfrac{\text{Number of even numbered term}}{2} × \left(\text{First term + Last term}\right)\\ = \dfrac{n}{2 × 2} × \left((a + d) + (a + (n - 1)d)\right)\\ = \dfrac{n}{4} × \left(2a + nd\right)$
Same reply goes for odd numbered terms.

Yes numbered of terms is $\dfrac{n}{2}$ but we have to divide by $2$ again as in the formula. You didnt do that. Hope this satisfies... :)

Ashish Menon - 5 years ago

Again going by your formula:- For even numbered terms $n$ in all places would be substituted by $\dfrac{n}{2}$ . $a$ would be substituted by $a + d$ . And $d$ would be substituted by $2d$ . Be careful with that one.

Ashish Menon - 5 years ago

Just arithmetic

The answer is 1002.

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