Think of the generalized form of n th term
We have 2 arithmetic progressions such that for all positive integers n , the ratio of the sum of first n terms is 7 n + 1 : 4 n + 2 7 .
If the ratio of the 11th terms of the AP can be expressed as p : q , where p and q are coprime positive integers, then find p + q .
The answer is 7.
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2 solutions
It wasn't clear to me that the ratio was for all n , instead of just for 1 value of n . Can you edit the problem for clarity?
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Thanks for the correction. I have edited the problem.
I have the following concerns with your solution:
- Shouldn't the sum be S n = 2 n ( 2 a + ( n − 1 ) d ?
- What does x − 1 = n − 1 mean? Didn't you originally set n = 2 x − 1 ?
- Assuming that such AP's exist, I believe that the ratio of the 11th term is 1 1 9 1 6 2 , where we substitue n = 2 3
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oh...i am sorry.I edited my solution.had written it in a hurry. But i am not satisfied with your 3rd comment.Can u explain how it works?
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Nope, your solution still doesn't make sense to someone who doesn't know how to solve this problem, because you provide no explanation of the steps / ideas.
E.g. When you say that the fraction is b 1 1 a 1 1 , are you hoping that it is equal to that ratio or are you claiming that it must be?
3rd comment - I made a typo, it should be n = 2 1 . It does yield 1 1 1 1 4 8 .
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@Calvin Lin – well we can clearly see that a 1 + 1 0 d 1 = a 1 1 and b 1 + 1 0 d 2 = b 1 1 .
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@Ayush G Rai – Because I know the solution, I can fill in the gaps in my head. However, for the person who doesn't know how to solve the problem, you are confusing them even further. You need to verbalize to the audience to explain what you are doing, especially at the crucial steps. Otherwise, they have to invent explanations for themselves, and they do not know for certain what you are thinking.
Some things that could improve the solution:
- Be explicit and state "Let the first AP have initial term a 1 and difference d 1 . Let the second AP have initial term b 1 and difference d 2 . (At this point in time, I'm confused why the initial terms are not a 1 and a 2 .)
- Explain that "Observe that if we set x = 1 1 , then this fraction will be equal to b 1 + ( 1 1 − 1 ) d 2 a 1 + ( 1 1 − 1 ) d 1 = b 1 1 a 1 1 ." By leaving it unstated, it is not clear to the audience what you are doing and why.
Note: The solution isn't to convince me that you have the right approach. The solution is to explain to others who could not solve the problem, how they can approach it and understand it.
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@Calvin Lin – well if we consider a 2 as the initial term of the 2nd AP then it would mean as the second term of the 1st AP. Thats why i didn't mention it.Well i will surely write all that u mentioned in the solution.
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@Ayush G Rai – Yes, so just say "initial terms a, b ". Because of the way you introduced the variables, the indices are interpreted to indicate "first progression, second progression".
Ayush your method is tedious , very tedious
this problem can be done by a simpler method
wtf is this?
Recall that if we know the sum of terms in an AP, then an easy way to find the nth term is to use:
Hence, the ratio of the 11th term, is going to be equal to the ratio of the sum of first 2 × 1 1 − 1 = 2 1 terms. By the condition, this is
4 × 2 1 + 2 7 7 × 2 1 + 1 = 1 1 1 1 4 8 = 3 4
Note: We have not established if such an AP exist. Does it?