Arithmetic Progressions (A-P)

[Achintya]

Q>The sum of n terms of two arithmetic progressions are in the ratio (3n+8): (7n + 15). Find the ratio of their 12th terms.

Note : I am already aware of the method wherein we subsitute 'n=23'.

But i have done this question with a different method and I am getting a wrong answer , please help me find the error. ( image ATTACHED)
Why cant we take the ratio of sum of 'n' terms as the absolute value of the sum of 'n' terms?
Where are exactly things going wrong???

Your help will be highly appreciated ???

 

[blamocur]

Whey you write [imath]S^1_n = x(3n+8)[/imath] and [imath]S^2_n = x(7n+15)[/imath] you are assuming that [imath]x[/imath] does not depend on [imath]n[/imath], which is not necessarily true.

 

[Achintya]

Whey you write [imath]S^1_n = x(3n+8)[/imath] and [imath]S^2_n = x(7n+15)[/imath] you are assuming that [imath]x[/imath] does not depend on [imath]n[/imath], which is not necessarily true.

Sir thank you so much, u r absolutely right ! Now i understand where things were going wrong.

If you don't mind i have one last question of the same type but slightly different ( Image attached below), it would be so great if you could identify the error in that too sir ???
Thanks in advance sir for your time ?

 

[blamocur]

Sir thank you so much, u r absolutely right ! Now i understand where things were going wrong.

If you don't mind i have one last question of the same type but slightly different ( Image attached below), it would be so great if you could identify the error in that too sir ???
Thanks in advance sir for your time ?

I don't understand what is in those pictures. Which problem are you solving?

 

[Achintya]

I don't understand what is in those pictures. Which problem are you solving?

Sir I am attaching the screenshot of the question here :

 

[blamocur]

I cannot see any problems in your pictures. At least I get the same answer of 7/97.

But what I've noticed that we don't use the fact that [imath]{a_k}[/imath] is an arithmetic progression. In fact, I do not believe that any A.P. can have the described property for arbitrary [imath]p,q[/imath]. Indeed, if you use [imath]p=1[/imath] you'll get [imath]S_q = a_1q^3[/imath], which is impossible since A.P. sums growth is of the quadratic order, not cubic. But if the statement is for some [imath]p,q[/imath], then I believe the problem is under-defined.

Where does this problem come from?

 

[Achintya]

Where does this problem come from?

Sir this problem was asked in the JOINT ENTRANCE EXAMINATION MAINS (JEE MAINS - Competitive exam in India for btech admissions).

If we solve this question with this method (Image attached below): then we do get the correct expected answer but i was just curious as to what is wrong with my method ....

 

[blamocur]

Your second solution is as correct as the first one. The fact that you get two different answers while using clean arguments tells me that something is wrong with the problem statement. Indeed, an arithmetic progression with this properties cannot exist:
[math]\frac{a_1+a_2}{a_1} = \frac{2^3}{1^3} = 8 \longrightarrow a_2 = 7a_1[/math][math]\frac{a_1+a_2+a_3}{a_1} = \frac{8a_1 + a_3}{a_1} = \frac{3^3}{1^3} =27 \longrightarrow a_3 = 19 a_1[/math]But [imath]\{a_1, 7a_1, 19a_1\}[/imath] cannot be consecutive items from an arithmetic progression unless [imath]a_1 = 0[/imath], in which case the answer to the problem is undefined.

I would point this out to whoever is in charge of this exam.

 

[Achintya]

Your second solution is as correct as the first one. The fact that you get two different answers while using clean arguments tells me that something is wrong with the problem statement. Indeed, an arithmetic progression with this properties cannot exist:
[math]\frac{a_1+a_2}{a_1} = \frac{2^3}{1^3} = 8 \longrightarrow a_2 = 7a_1[/math][math]\frac{a_1+a_2+a_3}{a_1} = \frac{8a_1 + a_3}{a_1} = \frac{3^3}{1^3} =27 \longrightarrow a_3 = 19 a_1[/math]But [imath]\{a_1, 7a_1, 19a_1\}[/imath] cannot be consecutive items from an arithmetic progression unless [imath]a_1 = 0[/imath], in which case the answer to the problem is undefined.

I would point this out to whoever is in charge of this exam.

You are absolutely right sir, I am so grateful to you sir for your help and your precious time.?
Once again thank you so much. Also I feel so lucky to have such a great community where students like me can connect with such knowledgeable and humble people like you .
Thanks a Ton sir !!!
Finally I can sleep peacefully???