As we continue through the End of Chapter exercises, we encounter a specialized type of problem: finding unknown terms when you only know their total sum and product.
If we assume the unknown terms are \(a\), \(a+d\), and \(a+2d\), the algebra gets incredibly messy. But if we use symmetric variables like \((a-d)\), \(a\), and \((a+d)\), adding them together instantly eliminates \(d\), allowing us to find the middle term immediately!
If we assume the unknown terms are \(a\), \(a+d\), and \(a+2d\), the algebra gets incredibly messy. But if we use symmetric variables like \((a-d)\), \(a\), and \((a+d)\), adding them together instantly eliminates \(d\), allowing us to find the middle term immediately!
1 Finding the Middle Term
2 Angles of a Triangle in AP
★ Complex Four-Term Ratios
Select the correct option:
When solving for three unknown consecutive terms in an AP, what is the best symmetrical algebraic assumption to use?
Concept:
1
In algebra, creating equations that allow variables to cancel each other out simplifies the problem instantly. Symmetric variables centered around \(a\) accomplish this.
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Hint:
1
Which option features a negative \(d\) and a positive \(d\) that sum to zero?
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Mistakes to Avoid:
!
Refusing to use symmetric terms: You CAN solve problems with \(a, a+d, a+2d\), but it forces you into difficult simultaneous equations involving both addition and multiplication simultaneously.
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Solution:
The optimal assumption is \(a-d, a, a+d\).
Their sum is \((a-d) + a + (a+d) = 3a\).
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Select the correct option:
The sum of 3 consecutive terms in an AP is 30. What is the value of the middle term?
Concept:
1
Because \((a-d) + a + (a+d) = 3a\), the sum of any three consecutive terms is always exactly three times the middle term.
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Hint:
1
Set \(3a = 30\).
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Mistakes to Avoid:
!
Assuming \(d=0\): Just because the middle term is 10, it does not mean the terms are 10, 10, 10. They could be 0, 10, 20 or 9, 10, 11. The middle term is absolute; the others depend on \(d\).
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Solution:
\(3a = 30 \implies a = 10\).
The middle term is \(a\), so the answer is 10.
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Select the correct option:
When solving for four unknown consecutive terms in an AP, which symmetrical algebraic assumption ensures that the common difference \(d\) cancels out when summed?
Concept:
1
To cancel out variables for an EVEN number of terms, we must split the gap around zero.
2
This forces the common difference of the assumed sequence to be \(2d\) instead of \(d\).
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Hint:
1
Test Option C: \((a+d) - (a-d) = 2d\). \((a+3d) - (a+d) = 2d\). The gap is perfectly constant!
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Mistakes to Avoid:
!
Choosing Option B: Many students mistakenly choose Option B because the numbers look consecutive (\(1, 2\)). But \((a+d) - (a-d) = 2d\), while \((a+2d) - (a+d) = d\). The gap changes, so it is NOT an Arithmetic Progression!
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Solution:
The correct assumption for four terms is \(a-3d, a-d, a+d, a+3d\).
Sum = \(4a\).
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Select the correct option:
If the sum of 4 consecutive terms in an AP is 20, what is the sum of the two middle terms?
Concept:
1
Because the terms are symmetrical, the sum of the inner pair and the sum of the outer pair are perfectly equal!
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Hint:
1
Sum of 4 terms = \(4a = 20 \implies a=5\).
2
Sum of middle two = \((a-d) + (a+d) = 2a\). Calculate \(2(5)\).
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Mistakes to Avoid:
!
Stopping at \(a\): If you calculate \(a=5\) and select Option A, you didn't answer the question! The question asks for the sum of the two middle terms.
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Solution:
Total sum: \(4a = 20 \implies a = 5\).
Middle terms sum: \((a-d) + (a+d) = 2a\).
\(2a = 2(5) = 10\).
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Select the correct option:
The sum of three consecutive terms in an AP is 15. If the first of these three terms is 2, what is the value of the third term?
Concept:
1
Sum of 3 terms = \(3a\). Find \(a\) (the middle term).
2
Once you have the first term and the middle term, you can instantly find \(d\).
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Hint:
1
\(3a = 15 \implies a = 5\). The middle term is 5.
2
The sequence looks like: \(2, 5, t_3\). What is the gap?
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Mistakes to Avoid:
!
Confusing \(a\) notation: In the symmetric assumption \(a-d, a, a+d\), the variable \(a\) represents the MIDDLE term, not the first term!
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Solution:
\(3a = 15 \implies a = 5\) (This is the middle term).
First term (\(a-d\)) = 2.
Therefore, \(5 - d = 2 \implies d = 3\).
Third term (\(a+d\)) = \(5 + 3 = 8\).
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