Section 15, easy algebra
Progressions (AP & GP)
Progression questions are mostly formula substitution: the nth term and sum of an AP, the sum and infinite sum of a GP. CAT dresses them up as savings plans, rows of seats, bouncing balls and “how many numbers between A and B”.
How CAT asks it
Terms or sums at two positions are given and you find others; you count or sum the numbers in a range with a divisibility property; you find the common terms of two APs; infinite GP sums (bouncing balls); word problems with steady increases; Sₙ given as a formula in n.
Concepts you must know
- AP: a, a + d, a + 2d, … with nth term Tₙ = a + (n − 1)d.
- AP sum = n × (first + last)/2, and the average of an AP = (first + last)/2.
- Three terms in an AP: write them as a − d, a, a + d (they sum to 3a).
- Number of terms from p to q with gap d = (q − p)/d + 1.
- GP: a, ar, ar², … with Tₙ = arⁿ⁻¹ and Sₙ = a(rⁿ − 1)/(r − 1).
- The common terms of two APs form an AP whose gap is the LCM of the two gaps.
- If Sₙ = An² + Bn, the series is an AP with d = 2A and first term A + B; any term is Tₙ = Sₙ − Sₙ₋₁.
- 1 + 2 + … + n = n(n + 1)/2; the sum of the first n squares is n(n + 1)(2n + 1)/6; of the first n cubes, [n(n + 1)/2]²; of the first n odd numbers, n².
- The AP sum and the infinite GP sum:
Thought process
- Decide AP (constant difference) or GP (constant ratio).
- Write a and d (or r). Two given facts give two equations.
- For counting in a range, find the first and last valid terms, then count.
- For bouncing balls, count the first drop once and every rebound height twice (up and down).
Patterns CAT repeats
Numbered so your error log can point at them: “Pattern 2”.
- 1Two terms given: find d and a, then any term or sum.
- 2Counting or summing numbers in a range divisible by k.
- 3Common terms of two APs.
- 4Infinite GP: bouncing balls, repeating decimals.
- 5Steady increases in word problems: savings, rows of seats.
- 6Sₙ given as a formula in n.
- 7Average of AP terms (overlaps with Section 9).
Worked examples
Attempt each one before opening its solution, then compare methods.
Example 1
Find the sum of all two-digit multiples of 3.
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They run from 12 to 99: (99 − 12)/3 + 1 = 30 terms. Sum = 30 × (12 + 99)/2. Answer: 1,665.
Example 2
The 5th term of an AP is 19 and the 12th is 47. Find the sum of the first 20 terms.
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7d = 28, so d = 4 and a = 19 − 16 = 3. S₂₀ = 10 × (2 × 3 + 19 × 4) = 10 × 82. Answer: 820.
Example 3
How many terms are common to 4, 9, 14, …, 199 and 3, 10, 17, …, 220?
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The first common term is 24 (in the second list: 3, 10, 17, 24). Common terms step by LCM(5, 7) = 35: 24, 59, 94, 129, 164, 199. The next, 234, passes 199. Answer: 6.
Example 4
A ball dropped from 20 m rebounds to 3/4 of its previous height every time. What total distance does it travel before stopping?
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20 down, then each rebound height up and down: 20 + 2 × [15/(1 − 3/4)] = 20 + 2 × 60. Answer: 140 m.
Example 5
Riya saves ₹500 in the first month and ₹100 more each month after that. After how many months will her total savings reach ₹11,000?
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Sₙ = n/2 × [1,000 + 100(n − 1)] = 50n² + 450n = 11,000, so n² + 9n − 220 = 0 = (n + 20)(n − 11). Check: 11 × (500 + 1,500)/2 = 11,000. Answer: 11 months.
Example 6
The sum of the first n terms of an AP is 3n² + 5n. Find the 10th term.
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T₁₀ = S₁₀ − S₉ = 350 − 288. Check with d = 6 and a = 8: 8 + 9 × 6 = 62. Answer: 62.
Traps
- Off-by-one errors when counting terms: it is (last − first)/d + 1.
- Using the infinite GP formula when |r| ≥ 1.
- Counting a bouncing ball’s first drop twice, or its rebounds once.
Practice set
Timed, about 2.5 minutes per question. Log every miss in the error log.
- 1
How many three-digit numbers are divisible by 7?
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128: from 105 (7 × 15) to 994 (7 × 142).
- 2
The 3rd term of an AP is 11 and the 7th is 27. Find the 15th term.
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59: d = 4 and a = 3.
- 3
Find the sum of the infinite GP 12 + 6 + 3 + …
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24: 12 ÷ (1 − 1/2).
- 4
Three numbers in an AP add to 24 and multiply to 440. Find them.
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5, 8 and 11: the middle term is 24/3 = 8, and 8(64 − d²) = 440 gives d = 3.
- 5
A theatre has 20 seats in the first row and 2 more in each row behind. How many seats are there in 15 rows?
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510: the last row has 48 seats, and 15 × (20 + 48)/2 = 510.
- 6
Find the sum of the first 20 odd numbers.
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400: the first n odd numbers add to n².