Solve the problem involving arithmetic progression

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SUMMARY

The discussion centers on clarifying the interpretation of the arithmetic progression problem from a past exam paper (specimen paper 2020, code 0606/01). Participants emphasize the importance of correctly defining the first term and the number of terms in the sequence, noting that the first term is given as 5 (i.e., n = 1). The confusion arises from the wording of the question, which lacks clarity in linking the number of terms to the variable n. A better-worded example from the June 2022 paper is also referenced, highlighting the need for precise language in mathematical problems.

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  • Knowledge of the conventions for defining terms in sequences
  • Experience with interpreting exam questions in mathematics
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  • Review the Cambridge IGCSE Mathematics syllabus for clarity on term definitions
  • Study the derivation of the sum of an arithmetic progression
  • Analyze past exam papers for common wording issues in mathematics questions
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chwala
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Homework Statement
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A.P
I posted this to clarify on the highlighted part- english problem for me.

First less than -200 means what?

1728785903791.png


Otherwise, the steps to solution are clear... cheers
 
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It means, the smallest ##n## for which the sum is less than -200.
 
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Hill said:
It means, the smallest ##n## for which the sum is less than -200.

The "number of terms" will be n + 1 rather than n if you start from n = 0, as would be usual.
 
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pasmith said:
The "number of terms" will be n + 1 rather than n if you start from n = 0, as would be usual.
But here it is stated that the first term (i.e. n = 1) is 5.
chwala said:
I posted this to clarify on the highlighted part- english problem for me.
The problem is with their English, not yours - they talk about "the number of terms" and they talk about "##n##" but they do not link the two. It should read "Find the number of terms ##n## such that..." or "Find ## n ## such that...".
 
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pbuk said:
But here it is stated that the first term (i.e. n = 1) is 5.

I don't agree.

It is natural to express a term of an arithmetic progression as a_n = c + dn with a_0 = c and not as a_n = c + d(n-1) with a_1 = c. In either case, the first term of the sequence is c.
 
pasmith said:
I don't agree.

I think we are splitting hairs about just how badly worded a badly worded question is. What did they really mean by "the first term is 5"? Does this imply that the first term is ## t_1 = 5 ##? If they had intended this to mean ## t_0 = 5 ## then would they have said "the zero'th term"?

Who knows, they don't even tell you that ## n ## is the number of terms so whether this starts at 0 or 1 is secondary.
 
The source of the paper is a past exam international paper. May be confusing to many students across the world.
 
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chwala said:
The source of the paper is a past exam international paper. May be confusing to many students across the world.
Indeed. Can you provide a link, or failing that state the exam board?
 
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chwala said:
The source of the paper is a past exam international paper. May be confusing to many students across the world.
Sorry, i just checked it is from the specimen paper of 2020 - code 0606/01. Most probably, this was corrected in subsequent papers i think...
 
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chwala said:
Sorry, i just checked it is from the specimen paper of 2020 - code 0606/01. Most probably, this was corrected in subsequent papers i think...
Ah, I see. Here is a link: https://www.cambridgeinternational....gcse-mathematics-additional-0606/past-papers/

In the June 2022 paper 1 there was a similar question which was indeed better worded:

7 (a) The first three terms of an arithmetic progression are ## \operatorname{lg} 3, 3 \operatorname{lg} 3, 5 \operatorname{lg} 3 ##. Given that the sum to ## n ## terms of this progression can be written as ## 256 \operatorname{lg} 81 ##, find the value of ## n ##. [5]​

Note also that in this syllabus (as I believe is the case for all GCSE and IGCSE syllabi), the convention is that the first term of a series is ## a_1 ##. This is also implied in the "Mathematical Formulae" section in the front of the paper:
1728988502560.png
 
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