This question is just nuts. Either $\frac {t_3} {t_1} \le 2$ and $S = \mathbb Z$ or $\frac {t_3} {t_1} > 2$ and $S =\emptyset$. If they are arguing that the $x$ must be an $x$ so that the the first conditional holds then they can't in any way claim an "or" as the first condition need not hold.

Wait I'm confused... How is S=Z if t3/t1<=2? We must satisfy t3/t1<=2 and extract the integer solutions... That's what I understood from S...

@fleablood why is S=Z if t3/t1<=2 ... Only 5 integers satisfy this inequality (as provided in their solution)

I'm with @fleablood. Was this question asked originally in English? I feel it is more a question of interpretation than of mathematics...

Yes yes... I'm confused. Please explain clearly why it's wrong so I can argue

I argued that: so your question is saying either that AP is true or S is true but sir said no: or need not mean you have to choose only one. You can chose both also. He said a+b= true means either a is true or b is true or both are true... and he said indepently find the values of x that satisfy the AP and then find the set S... Im confused

The set $\{x|$the sky is blue$, x\in \mathbb Z\}=\mathbb Z$ because for every integer the sky is blue. $\{x|$ elephants are pink$, x\in \mathbb Z\}=\emptyset$ because there is no integer for which elephants are pink. So $\{x|\frac {t_3}{t_1}\le 2, x\in \mathbb Z\}$ will either be all integers or no integers depending on whether $\frac {t_3}{t_1}\le 2$. Unless you are claiming that $t_3$ and $t_1$ are somehow functions based on $x$. But they CANT be because the first premise of the OR doesn't have to be true.And if it is true the second half doesn't have to be true. The question is NUTS.

"OR $S = \left\{x\Bigg| \dfrac{t_3}{t_1}\leq2 : x \in \mathbb{Z} \right\}\implies \frac{|2x^2 - ||x|+5||}{|x^2 - |4x+3||}\leq2, x \in \mathbb{Z}$" This is false. $\dfrac{t_3}{t_1}\ne \frac{|2x^2 - ||x|+5||}{|x^2 - |4x+3||}$. We have no idea what $\dfrac{t_3}{t_1}$ is because the first condition need not be true. So we don't know ANYTHING about the terms of the AP.

@fleablood Ok now I understand... How can we substitute those values for t1 and t3 if the first condition(AP) need not be true... But one last thing... sir said: "read the last two words of the question, it's written can be, so he said there is a possibility that it is true, and we can substitute... But that leads to another confusion. If we do substitute, arn't we restricting x to be only those 2 integer values of x(namely: 2,5/3)... Please correct if the word can be is being misused...

@fleablood: he said you are right it need not be true: but he said the words can be are key to the qstn... :/, Please clear this up. Thanks

@fleablood even if this word can be means that we can take t3/t1 be those functions of x, isn't the solution still wrong? because to find the 3|S| terms of the AP, arn't we restricting the values of x to be only those values satisfying the AP and in that case S ={2}, not those other 4 integers as those values of x dont satisfy the AP? Again if the word can be is being misused, please correct me...

The question is not answerable since we don't know what $t_1,t_3$ are, but If "or" is meant to be "with", and $t_1=|x^2 - |4x+3||,t_3=|2x^2 - ||x|+5||$, then the question should be answerable since it is equivalent to "If $x\in\{\frac 53,2\}$ with $S =\{-2,0,1,2,3\}$, then the sum of the first $3|S|$ terms of the decreasing AP can be...".

@mathlove... If or is replaced with 'with' then S won't be those 5 terms because x is restricted to be only 5/3 and 2, so shouldn't S={2}?? If I'm wrong, correct me

@Sid: No, it shouldn't. $S$ is defined as $\left\{x \in \mathbb{Z}: \dfrac{t_3}{t_1}\leq2 \right\}$, not as $\left\{x \in \mathbb{Z}: \text{$t_1, 4, t_3$ are the first three terms of a decreasing AP, and}\ \dfrac{t_3}{t_1}\leq2 \right\}$.

@mathlove Consider this simpler statement: x = 1 with S = {x:x<2}, will S = (-$\infty$,2) or will S be solely 1? Because x is already locked to be = 1. Am I wrong?

@Sid : $S$ is defined as the set of integers $x$ satisfying $\frac{|2x^2 - ||x|+5||}{|x^2 - |4x+3||}\leq2$. If you really understand this definition, you should be able to understand that one can write $S = \left\{y \in \mathbb{Z}: \frac{|2y^2 - ||y|+5||}{|y^2 - |4y+3||}\leq2 \right\}$.

Please rewrite the question. Why are you saying $x \in \mathbb{Q}$ if you then just restrict to $\mathbb{Z}$? I'm not exactly sure what $S$ is.

@mathworker21... Read the comments completely... I had the same doubt as you about S... But look what mathlove wrote just about your comment.

@Sid No. You should fix the question. I'm not reading every single comment because you want to keep your question written horrendously.

@Sid No. Your question is written terribly. I don't need to read comments to know that. However, I did read that comment, and, no surprise, nothing changed; fix ur question

Uhm the question doesn't look like anything how I'd written it... Someone has already eddited it. Check the history

And dude. I asked @mathlove wouldn't the values of x in the AP affect the set S, he said no my mathematical reasoning was wrong. Just take some time to read it and you'll get the answer. If the question is written like this in the test what give me the right to change it???

$15$ doesn't divide $-200$.

Ya that was my doubt... Does the divisor have to divide all the possible values of the sum??

@mathlove is there still something wrong with the wording of the question? You yourself said if the or is switched to with then the qstn makes sense... who is this person saying like this??

@Sid : I've just posted an answer. See the answer.

$25$ also divides two of the three possible values, so if $15$ were a correct answer, then all the answers would be correct (which would make it a silly question, no?). When it says, "the sum is divisible by ..." it means, in all cases the sum is divisible by ....