Progression (AP & GP)

Look at the AP 5, 8, 11, 14, … To reach the 2nd term you add d once. To reach the 3rd term you add d twice, and to reach the 10th term you add d nine times. So the nth term is the first term plus (n − 1) jumps of d:

T_n = a + (n − 1)d

This (n − 1) is the heart of the formula. The first term is already standing at position 1, so you need only (n − 1) more jumps to reach position n. For 5, 8, 11, …: T₂₀ = 5 + 19 × 3 = 62.

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Number of terms. If the first term a, the last term l and d are known, turn the same formula around: l = a + (n − 1)d, so

n = l − ad + 1

How many terms are there in 12, 17, 22, …, 132? n = 132 − 125 + 1 = 24 + 1 = 25. The "+1" is there because both ends are counted. It is the same counting you do when you find how many numbers from 12 to 132 leave the same remainder on division by 5.

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Term from the end. Stand at the last term l and walk backwards; every step subtracts d. So the nth term from the end = l − (n − 1)d.

Gap between two terms. T_m − T_n = (m − n)d, and a is not needed at all. If T₉ = 40 and T₄ = 20, then 5d = 20, so d = 4, and a = 20 − 3 × 4 = 8.

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When some term is zero. Suppose Radhika is told that the 12th term of an AP is zero. Then a + 11d = 0, so a = −11d, and every term can now be written in d alone: T₅ = −11d + 4d = −7d and T₈ = −11d + 7d = −4d. So 4 × T₅ = −28d = 7 × T₈. The d cancels and a clean number is left, which is how such "how many times" questions are built.

The most useful property of an AP is this: every term is the average of its two neighbours. In 10, 16, 22 the middle term 16 = 10 + 222. Why? Because 16 is d more than 10 and d less than 22, so it sits exactly in the middle.

So three numbers a, b, c are in AP if and only if

2b = a + c

The middle number b is called the arithmetic mean (AM) of a and c: AM = a + c2.

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The average of a whole AP. The same idea works for the full progression. Pair the first and last terms, the second and second-last, and so on; every pair has the same sum (a + l). So

Average of an AP = first term + last term2, which is also the middle term when the number of terms is odd.

Radhika writes the AP 9, 13, 17, 21, 25. Its average = 9 + 252 = 17, which is also the middle term. Check: 9 + 13 + 17 + 21 + 25 = 85, and 85 ÷ 5 = 17. With an even number of terms, say 9, 13, 17, 21, the average is 9 + 212 = 15, the average of the two middle terms.

Inserting means. If n numbers are inserted between a and b so that the whole list is an AP, there are (n + 1) gaps between a and b, so d = b − an + 1. Insert 3 numbers between 6 and 26: d = 204 = 5, giving 6, 11, 16, 21, 26.

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Selection of terms. When a question says "three numbers in AP have sum …", don't take them as a, a + d, a + 2d. Take them symmetrically, so that d cancels in the sum:

• 3 terms: (a − d), a, (a + d); sum = 3a
• 4 terms: (a − 3d), (a − d), (a + d), (a + 3d); sum = 4a, common difference 2d
• 5 terms: (a − 2d), (a − d), a, (a + d), (a + 2d); sum = 5a

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AP in geometry. If the radii (or diameters) of circles are in AP, their circumferences are also in AP, because circumference = 2πr just multiplies every term by the same number 2π. Multiplying or dividing every term of an AP by the same non-zero number keeps it an AP, and so does adding the same number to every term.

You already know the method from the start of the chapter: write the sum forward and backward and add. Every column gives (first + last), and there are n columns. So for any AP:

S_n = n2(a + l)

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Since l = a + (n − 1)d, the same thing can be written without l:

Use the first form when the last term is known, which is the case in most series questions, and the second form when d is given but the last term is not.

Now read the first formula once more. a + l2 is the average of the AP, so

Sum of an AP = number of terms × average of the first and last term

This one line solves almost every series question. Take 51 + 52 + … + 90. Number of terms = 90 − 51 + 1 = 40, average = 51 + 902 = 70.5, so the sum = 40 × 70.5 = 2820.

Series of multiples. 4 + 8 + 12 + … + 160 is just 4 × (1 + 2 + … + 40) = 4 × 40 × 412 = 3280. Take the common factor out, and you are left with a sum of natural numbers.

Sum from m to n. Find 1 + 2 + … + n and subtract 1 + 2 + … + (m − 1). For 21 + 22 + … + 50: 50 × 512 − 20 × 212 = 1275 − 210 = 1065. (Check: 30 terms × average 35.5 = 1065.)

Numbers divisible by k in a range. First find the first and last multiple of k in the range, then the number of terms n = last − firstk + 1, then sum = n × first + last2. For two-digit numbers divisible by 6: first 12, last 96, n = 846 + 1 = 15, sum = 15 × 54 = 810.

The mountain series. 1 + 2 + 3 + … + (n − 1) + n + (n − 1) + … + 2 + 1 = n². Why? The left part 1 + … + (n − 1) and the right part (n − 1) + … + 1 are the same, so the total = 2 × (n − 1)n2 + n = n² − n + n = n². For example, 1 + 2 + … + 9 + 10 + 9 + … + 1 = 10² = 100.

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Getting a term back from the sum. T_n = S_n − S_{n − 1}. The sum of n terms minus the sum of (n − 1) terms leaves only the nth term. Also, if S_n is given in the form An² + Bn, the series is an AP with common difference 2A.

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When the question gives conditions on sums. Sometimes S_n itself is given and you must get back to a and d. Write S_n = n2[2a + (n − 1)d] for each given n, cancel the number outside, and two small equations in a and d are left. If S₁₀ = 175 and S₂₀ = 650, we get 2a + 9d = 35 and 2a + 19d = 65. Subtract: 10d = 30, so d = 3 and a = 4.

Two results save a lot of time here.

1. If S_m = S_n for m ≠ n, then S_m+n = 0. Write S_m − S_n = 0 as (m − n)[2a + (m + n − 1)d] = 0. Since m ≠ n, the bracket must be zero, and that bracket is exactly the one inside S_m+n. So if the sum of the first 12 terms equals the sum of the first 20 terms, the sum of the first 32 terms is 0.

2. Ratio of the sums of two APs. If S_n : S′_n = f(n) : g(n), then T_n : T′_n = f(2n − 1) : g(2n − 1). The reason is simple: S_2n−1 = (2n − 1) × T_n, because T_n is the middle term of the first 2n − 1 terms. So replace n by (2n − 1) to get the ratio of the nth terms.

Two values of n. When d is negative, the sum rises and then falls, so the same sum can appear twice. For 24, 20, 16, … a sum of 72 gives n² − 13n + 36 = 0, so n = 4 or n = 9. The terms from the 5th to the 9th (8, 4, 0, −4, −8) add up to zero, which is why both answers are correct.

Whenever something increases or decreases by the same amount every day, week, month or year, the story is an AP. Your job is to translate the story into a, d and n, and then decide one thing: does the question ask for one term (T_n) or for the total (S_n)?

Read the last line carefully:

• "What will be the salary in the 6th year?", "How much does he deposit in the 12th week?" → one term, use T_n = a + (n − 1)d.
• "What is the total earning in 6 years?", "How much has he saved by the end of 12 weeks?" → total, use S_n = n2[2a + (n − 1)d].

Radhika saves ₹30 in the first month, and every month she saves ₹15 more than the month before. In the 8th month she saves T₈ = 30 + 7 × 15 = ₹135. In the first 8 months together she saves S₈ = 82(30 + 135) = 4 × 165 = ₹660. The story is the same; the two questions need two different formulas.

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Speed increasing every hour. If a vehicle moves at speed a in the 1st hour, (a + d) in the 2nd hour and so on, the distance covered in each hour is a, a + d, a + 2d, … (distance in one hour = speed × 1 hour). So the total distance in n hours = S_n. When the distance is given and the time is asked, you get a quadratic in n. It is quicker to try the options: for each option n, work out n × first + last2 and see which one gives the distance.

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"Next 5 years" language. If a salary is ₹x in the first year and an increment comes in each of the next 5 years, the total period is 6 years (1 + 5). Count the terms from the story with care.

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In a GP you don't add, you multiply. Every term is the previous term × r. So the 2nd term is a × r, the 3rd is a × r × r, and the nth term has (n − 1) multiplications by r:

T_n = ar^{n − 1}

It is the same (n − 1) logic as in the AP; only the jump is now a multiplication. A plant in Radhika's garden has 3 leaves on day 1, and the number of leaves doubles every day. On day 7 it has 3 × 2⁶ = 3 × 64 = 192 leaves.

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Sum of n terms. For r ≠ 1:

S_n = a(rⁿ − 1)r − 1 (handy when r > 1)
S_n = a(1 − rⁿ)1 − r (the same formula, easier when r < 1)

Why does it work? Write S = a + ar + … + ar^{n − 1}. Multiply by r: rS = ar + ar² + … + arⁿ. Subtract, and all the middle terms cancel: rS − S = arⁿ − a. So S = a(rⁿ − 1)r − 1. If r = 1, every term is a and S_n = na.

For example, 5 + 15 + 45 + 135 + 405 has a = 5, r = 3, n = 5, so S = 5(3⁵ − 1)3 − 1 = 5 × 2422 = 605.

Infinite GP. When the ratio lies between −1 and 1 (−1 < r < 1, r ≠ 0), the terms keep shrinking towards zero and the sum of infinitely many terms settles at a fixed value:

S_∞ = a1 − r, only when |r| < 1

8 + 4 + 2 + 1 + … = 81 − 12 = 16. Recurring decimals are infinite GPs too: 0.4 = 0.4 + 0.04 + 0.004 + … = 0.41 − 0.1 = 49. If |r| ≥ 1, the infinite sum does not exist.

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Geometric mean (GM). Three non-zero numbers a, b, c are in GP if ba = cb, that is

b² = ac, and the middle term b = √ac is called the GM of a and c.

GM of 4 and 25 = √100 = 10, and 4, 10, 25 is a GP with r = 52. For three terms in GP, take them as ar, a, ar. Their product is a³, which makes product-based questions easy.

AM, GM and HM together. For two positive numbers a and b:

• AM = a + b2, GM = √ab, HM (harmonic mean) = 2aba + b
• AM ≥ GM ≥ HM (equal only when a = b), and GM² = AM × HM

For 4 and 25: AM = 14.5, GM = 10, HM = 20029. Check: 14.5 × 20029 = 100 = 10².

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Inserting geometric means. If n numbers are put between a and b so that the whole list is a GP, there are (n + 1) multiplications from a to b. So b = a × r^{n + 1}, which gives

r = (ba)^{1n + 1}

Insert 3 numbers between 3 and 48: r⁴ = 483 = 16, so r = 2 and the list is 3, 6, 12, 24, 48. The three inserted numbers are 6, 12, 24. A neat fact: the product of the n inserted means = (ab)ⁿ², because the means pair up end to end and every pair multiplies to ab. Here 6 × 12 × 24 = 1728 = (3 × 48)³² = 144³² = 12³.

Harmonic progression (HP). A sequence is an HP when the reciprocals of its terms form an AP. So 1, 13, 15, 17, … is an HP, because 1, 3, 5, 7, … is an AP. To work with an HP, turn every term upside down, solve the AP question, and turn the answer upside down again. The nth term of an HP is 1a + (n − 1)d, where a and d belong to the reciprocal AP. For 12, 6, 4, 3, … the reciprocals are 112, 16, 14, 13, which is an AP with d = 112, so the 8th term is 1112 + 7 × 112 = 128 = 32. The harmonic mean of a and b is the middle term of the HP a, HM, b, which is the HM = 2aba + b you already know.

Some series are neither AP nor GP, but each of the four families below has a ready formula. Learn them once and these questions become direct.

1. Sum of squares.
1² + 2² + … + n² = n(n + 1)(2n + 1)6
For n = 10: 10 × 11 × 216 = 385. Keep 385 in memory; it is a handy value for checking. Before multiplying, cancel the 6 against the factors on top, and the numbers stay small.

From m to n: subtract. 6² + 7² + … + 15² = (1² + … + 15²) − (1² + … + 5²) = 1240 − 55 = 1185. Subtract up to (m − 1), not up to m.

2. Sum of cubes.
1³ + 2³ + … + n³ = [n(n + 1)2]²
The sum of cubes is the square of the sum of natural numbers: 1³ + 2³ + … + 5³ = 15² = 225.

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3. Pattern series. When the differences are not constant, look at the differences of the differences. If those are constant, the nth term has the form An² + Bn + C, and often it is something simple like n² + k, k·n² or n(n + 1). Radhika is given 3, 6, 11, 18, … The differences are 3, 5, 7, going up by 2. The terms are 1 + 2, 4 + 2, 9 + 2, 16 + 2, that is n² + 2, so the next term = 25 + 2 = 27.

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4. Cancelling (telescoping) fractions. When every denominator is a product of two numbers that differ by the same amount d, split each fraction:

1k(k + d) = 1d(1k − 1k + d)

Everything in the middle then cancels, and only the first and the last pieces survive. For example, in 11 × 4 + 14 × 7 + 17 × 10 we have d = 3, so the sum = 13(1 − 14 + 14 − 17 + 17 − 110) = 13(1 − 110) = 13 × 910 = 310.

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So the working rule is sum = 1d × (1first number − 1last number), where "first number" is the first factor of the first denominator and "last number" is the second factor of the last denominator. When the denominators are not written as products (like 15, 35, 63, …), first break them into factors and check that the factors of neighbouring denominators overlap.

5. Numbers made of one repeated digit, and sums of digits. For a series like 9 + 99 + 999 + 9999, write every term as (10^k − 1): (10 − 1) + (100 − 1) + (1000 − 1) + (10000 − 1) = 11110 − 4 = 11106. For 6 + 66 + 666 + 6666, take 6 out: each term is 69 × (10^k − 1), so the sum = 69 × (11110 − 4) = 23 × 11106 = 7404.

Sum of the digits of all numbers from 1 to 99. Write the numbers as 00, 01, …, 99. Each of the two places carries every digit 0 to 9 exactly ten times, and 0 + 1 + … + 9 = 45. So the total = 2 × 10 × 45 = 900. For 1 to 999 the same idea gives 3 × 100 × 45 = 13500. For a range that stops in between, split it into full blocks: the digits of 1 to 50 add up to 45 (for 0 to 9) + 55 + 65 + 75 + 85 (for the blocks 10–19, 20–29, 30–39, 40–49) + 5 (for 50) = 330.

Alternating series. In 1² − 2² + 3² − 4² + … pair the terms: a² − (a + 1)² = −(2a + 1). The pairs −3, −7, −11, … are an AP, or more directly the sum of 1² − 2² + … + 19² − 20² equals −(1 + 2 + … + 20) = −210.

Whenever something changes by the same factor every time (it doubles every hour, it loses 10% of its value every year, a ball rebounds to 23 of its last height), the story is a GP. Your first job is to find the factor r. Your second job is to decide what the question wants: one term, a total, or a total that goes on for ever.

Growth and decay. Growth by p% in every period means r = 1 + p100. A fall by p% means r = 1 − p100. A machine worth ₹80,000 loses 10% of its value every year, so r = 0.9. Its value at the start is the first term, which means the value after n years is the (n + 1)th term: after 3 years the machine is worth 80,000 × 0.9³ = ₹58,320.

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The bouncing ball. A ball is dropped from a height H and after every fall it rebounds to a fraction r of the height it fell from. The heights are H, rH, r²H, … The path is a single fall of H, and after that every height is travelled twice, once up and once down:

Total distance = H + 2rH + 2r²H + … = H + 2rH1 − r = H × 1 + r1 − r

For H = 80 m and r = 35: total = 80 × 1.60.4 = 320 m. If the question stops at a particular bounce, add only the heights up to that bounce; do not use the infinite formula.

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Yearly deposits. Radhika puts ₹1,000 in a bank at the start of every year and the bank pays 10% compound interest. The first deposit earns interest for 3 years, the second for 2 years, the third for 1 year. At the end of 3 years she has 1,000 × (1.1³ + 1.1² + 1.1) = 1,000 × 3.641 = ₹3,641. The amounts form a GP, and the answer is its sum. Compound interest on a single sum is the same idea: P, P(1 + R100), P(1 + R100)², … are the amounts at the end of the 0th, 1st, 2nd year.

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Chain stories. "One person tells a secret to 3 people in the first hour, and each of them tells 3 new people in the next hour." The new people in each hour are 3, 9, 27, … Be clear whether the question asks for the people told in a particular hour (a term) or by the end of that hour (a sum).