We have finally reached the End of Chapter exercises! This is where we combine everything we have learned about Arithmetic Progressions.
These problems will not just ask you to plug numbers into a formula. They will test your logical reasoning, requiring you to set up simultaneous equations, count specific sets of numbers, and manipulate algebraic rules. Let's conquer the final gauntlet!
These problems will not just ask you to plug numbers into a formula. They will test your logical reasoning, requiring you to set up simultaneous equations, count specific sets of numbers, and manipulate algebraic rules. Let's conquer the final gauntlet!
1 Working Backwards in an AP
2 Counting Divisible Numbers
3 Simultaneous Term Equations
Select the correct option:
In an AP, if \(d = -4\), \(n = 7\), and the \(n^{th}\) term \(t_n = 4\), what is the value of the first term (\(a\))?
Concept:
1
You can solve the explicit formula \(t_n = a + (n-1)d\) for any missing variable if you know the other three.
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Hint:
1
Substitute the given values: \(4 = a + (7-1)(-4)\).
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Mistakes to Avoid:
!
Sign errors across the equals sign: When you get \(4 = a - 24\), you must add 24 to the left side. \(4 + 24\), not \(24 - 4\)!
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Solution:
\(t_n = a + (n-1)d\)
\(4 = a + (7-1)(-4)\)
\(4 = a + 6(-4)\)
\(4 = a - 24\)
\(a = 4 + 24 = 28\).
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Select the correct option:
What is the common difference (\(d\)) of an AP in which \(t_{18} - t_{14} = 32\)?
Concept:
1
The difference between any two terms in an AP is simply the number of jumps between them multiplied by \(d\).
2
\(t_y - t_x = (y - x)d\).
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Hint:
1
Expand \(t_{18}\) to \((a + 17d)\). Expand \(t_{14}\) to \((a + 13d)\).
2
Subtract them: \((a + 17d) - (a + 13d) = 4d\). So \(4d = 32\).
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Mistakes to Avoid:
!
Solving for 'a': You do not need to know the first term to find the difference between two arbitrary terms!
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Solution:
\(t_{18} - t_{14} = 32\)
\((a + 17d) - (a + 13d) = 32\)
\(4d = 32 \implies d = 8\).
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Select the correct option:
Two Arithmetic Progressions have the exact same common difference. The first term of one AP is 1, and the first term of the other AP is -8. What is the difference between their 4th terms?
Concept:
1
If two sequences have the same slope/growth rate (\(d\)), the distance between them is constant forever.
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Hint:
1
AP 1: \(1, 1+d, 1+2d, 1+3d\)
2
AP 2: \(-8, -8+d, -8+2d, -8+3d\)
3
Subtract AP 2's 4th term from AP 1's 4th term. The \(d\)'s cancel out!
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Mistakes to Avoid:
!
Assuming \(d\) is required: Because you are comparing two identical rates, the \(d\) variable cancels out mathematically. You do not need its value!
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Solution:
Difference = \(t_4(\text{AP1}) - t_4(\text{AP2})\)
Difference = \((1 + 3d) - (-8 + 3d)\)
Difference = \(1 + 3d + 8 - 3d\)
Difference = \(9\).
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Select the correct option:
Which term of the AP \(3, 15, 27, 39 \dots\) will be exactly 120 more than its 21st term?
Concept:
1
You can avoid massive calculations. If you need a value "120 more", simply divide 120 by the common difference \(d\) to see how many "jumps" forward that represents!
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Hint:
1
Calculate \(d = 15 - 3 = 12\).
2
How many jumps is 120? \(\frac{120}{12} = 10\) jumps.
3
Add 10 jumps to the 21st position.
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Mistakes to Avoid:
!
Calculating the massive values: You CAN find \(t_{21}\), add 120 to it, and then solve \(a + (n-1)d = \text{Target}\), but that takes forever and leads to arithmetic errors. Use relative logic!
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Solution:
\(d = 15 - 3 = 12\).
Target term = \(t_{21} + 120\).
Number of additional terms = \(\frac{120}{12} = 10\).
Therefore, the position is \(21 + 10 = 31\).
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Select the correct option:
If a sequence is written perfectly backward (e.g. \(100, 95, 90 \dots 5\)), what happens to the mathematical properties of the AP?
Concept:
1
Reversing an AP is a powerful trick. By reading it right-to-left, the end point is the new starting point, and increasing gaps become decreasing gaps.
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Hint:
1
If AP is \(5, 10, 15\dots\) then \(d = +5\).
2
If reversed to \(\dots15, 10, 5\) then \(d = -5\).
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Mistakes to Avoid:
!
Keeping \(d\) constant: If you reverse the sequence but keep \(d = +5\), the numbers will grow off into infinity rather than shrinking back down to the original start.
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Solution:
When reversing an AP, the last term \(L\) serves as the new first term \(a_{new}\), and the new common difference \(d_{new}\) is exactly \(-d_{old}\).
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★ Complex Variable Proofs
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