Chapter 6 of 9

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Telangana (TSBIE)Class 10 Mathematics★ The textbook's index allots this chapter 11 periods in January, and it runs from textbook page 129 to page 162 - 34 pages, the longest single chapter span in the first half of the book. That is five numbered exercises (6.1 with four questions, 6.2 with seventeen, 6.3 with fourteen, 6.4 with four, 6.5 with seven) plus an Optional Exercise of seven, twenty-two worked examples, five Do This boxes, three Try This boxes, two Think and Discuss boxes and one Activity. No marks distribution is printed anywhere in the volume, so no official weightage can be quoted; the period count is the only number the book itself gives.

Progressions

What separates a pattern that merely repeats from one that progresses, how the nth term and the sum of an arithmetic progression are built, the pairing trick Gauss used to add 1 to 100, and how a geometric progression differs by multiplying where an A.P. adds. Telangana SCERT Class 10 Mathematics Chapter 6.

⏱ 21 min readMedium difficulty✓ Free · TSBIE syllabus

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By the end of this chapter you'll be able to…

1Distinguish a repeating pattern from a progressive one, and say which of the chapter's opening lists are which
2Define an arithmetic progression and identify its first term a and common difference d
3Test whether a list is an A.P. by subtracting consecutive terms in the correct order, later minus earlier
4Write the terms of an A.P. given a and d, including the cases d negative and d zero
5Use aₙ = a + (n - 1)d to find any term, and to find n, a or d when the others are known
6Decide whether a given number belongs to an A.P., recognising that a fractional n means it does not
7Count how many numbers in a range satisfy a divisibility condition by treating them as an A.P.
8Find a term counted from the end of a finite A.P. by first finding the total number of terms
9Derive the sum formula by writing the sum forwards and backwards and adding, as Gauss did
10Apply Sₙ = n/2 [2a + (n - 1)d] and Sₙ = n/2 (a + l), choosing whichever the given data suits
11Interpret two roots of a quadratic in n, deciding whether both are admissible or one must be rejected
12Define a geometric progression, find its common ratio, and use aₙ = a r^(n-1) to find any term

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Why this chapter matters

This chapter teaches a distinction that sounds obvious and is not: the difference between a pattern that repeats and a pattern that progresses. A sunflower's petals repeat; a salary with a yearly increment progresses. Once that line is drawn, two operations account for almost every progressive pattern a student will meet - adding the same amount, or multiplying by the same amount - and each gets a formula for its nth term so that the hundredth item never has to be reached by counting. The sum formula matters even more, because it comes with a method rather than just a result: Gauss's trick of writing a sum forwards and backwards and adding is a genuine piece of mathematical thinking, reusable far beyond this chapter. Progressions also underpin compound interest, depreciation, population growth and instalment schemes, so this is where school arithmetic starts to describe money and growth honestly.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Number patterns and rules for continuing a sequence, from earlier classes

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Solving linear equations in one variable, needed to find n from the nth-term formula

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Solving quadratic equations by factorisation, needed for the sum questions in Exercise 6.3

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Laws of exponents, especially a^m divided by a^n, used throughout the geometric progression work

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Simple and compound interest, which supply the chapter's arithmetic and geometric examples respectively

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Area of a square and perimeter of a triangle, used in the midpoint-square and midpoint-triangle examples

Progressions

1. What This Chapter Covers

The chapter opens in nature: the petals of a sunflower, the cells of a honeycomb, the grains on a maize cob, the spirals on a pineapple and a pine cone. All of them show pattern. But the book immediately draws a distinction that the whole chapter rests on.

Those natural patterns repeat; they do not progress. The petals of a sunflower are equidistantly grown, and a honeycomb's identical hexagons are arranged symmetrically around each other. Nothing grows or shrinks from one to the next.

Compare four lists the book then gives. The last digits of 4, 4², 4³, 4⁴, ... run 4, 6, 4, 6, 4, 6 — a repetition. But Usha's salary of Rs 8000 with a Rs 500 annual increment runs 8000, 8500, 9000, and the rungs of a ladder shortening by 2 cm run 45, 43, 41, 39, 37, 35, 33, 31.

In those last two, the relationship between the numbers is constantly progressive. Each succeeding term is obtained by adding a fixed number to the one before. That is what this chapter is about, together with its multiplicative cousin.

The index allots this chapter 11 periods in January, and it runs from textbook page 129 to page 162 — the longest page span in the book so far, with five numbered exercises and an Optional Exercise.

2. Two Ways to Progress

Not every progressive pattern adds. In a savings scheme where the amount becomes 5/4 times itself every three years, an investment of Rs 8000 grows to 10000, then 12500, then 15625, then 19531.25. Nothing fixed is being added; something fixed is being multiplied.

The straight line on the left is not decoration. Because an arithmetic progression adds the same amount each step, its terms always fall on a straight line; a geometric progression curves away.

The chapter's History note records that by 400 BCE the Babylonians knew of arithmetic and geometric progressions, that they were known to early Greek writers, and that among Indian mathematicians Aryabhata was the first to give a formula for the sum of squares and cubes of natural numbers, with Brahmagupta, Mahavira and Bhaskara also considering such sums.

3. What an Arithmetic Progression Is

An arithmetic progression is a list of numbers in which each term, except the first, is obtained by adding a fixed number to the preceding term. That fixed number is the common difference, written d.

Writing the first term as a and the common difference as d, the A.P. is a, a + d, a + 2d, a + 3d, ... — its general form. Two numbers are enough to generate the whole list.

The book stresses that d can be anything. With a = 6 and d = 3 the A.P. is 6, 9, 12, 15; with a = 6 and d = −3 it is 6, 3, 0, −3; with a = 2 and d = 0 it is 2, 2, 2, 2. A constant list is a perfectly good A.P.

To test a list, subtract consecutive terms. If a₂ − a₁ = a₃ − a₂ = ... the list is an A.P. The book adds a warning about direction: to find d in 6, 3, 0, −3 you subtract 6 from 3, not 3 from 6. Always take the later term minus the earlier one, even when the result is negative.

One example is worth holding on to. The list 1, 1, 2, 3, 5, ... — the rabbit-pairs sequence from the introduction — is not an A.P., because the gaps 0, 1, 1, 2 are not constant. A famous pattern need not be an arithmetic one.

An A.P. with a last term is finite; one that never ends is infinite. The heights 147, 148, ..., 157 and the cash prizes Rs 200 to Rs 750 are finite; 1, 2, 3, 4, ... is not.

The Try This box on page 132 is a good check of whether the definition has landed. Of 2, 3, 5, 7, 8, 10, 15 and 2, 5, 7, 10, 12, 15 and −1, −3, −5, −7, only the third is an A.P.; the first two look orderly but their gaps alternate rather than repeat.

A second box asks something more interesting: take any A.P., then add a fixed number to every term, subtract a fixed number from every term, or multiply every term by a fixed number, and check each result. All three stay arithmetic. Adding k to every term shifts a to a + k and leaves d alone; multiplying every term by k scales both a and d by k.

4. The nth Term

Usha's salary makes the point. Adding Rs 500 repeatedly to find her salary in the 25th year is possible but tedious. The book notices the shortcut in its own working: her 15th-year salary is 8000 plus 500 added fourteen times, and her 25th-year salary is 8000 plus 500 added twenty-four times.

That generalises at once. The second term is a + (2 − 1)d, the third is a + (3 − 1)d, the fourth is a + (4 − 1)d, so

aₙ = a + (n − 1)d

and aₙ is called the general term. If the A.P. has m terms, aₘ is the last term, often written l.

The "minus one" is where marks are lost. The first term needs no additions at all, so the nth term needs n − 1 of them.

Example-4 shows the formula answering two questions at once. In 21, 18, 15, ... the term equal to −81 is found by solving −81 = 21 + (n − 1)(−3), which gives n = 35. Asking instead which term is zero gives 21 + (n − 1)(−3) = 0 and n = 8.

Example-6 shows it answering a third kind of question. Is 301 a term of 5, 11, 17, 23, ...? Solving 301 = 5 + (n − 1)6 gives n = 302/6, which is 151/3. Since n must be a positive integer, 301 is not a term at all. A fractional n is the signal that a number is not in the list.

Example-7 turns the formula into a counting tool. The two-digit multiples of 3 are 12, 15, 18, ..., 99, an A.P. with a = 12 and d = 3, and solving 99 = 12 + (n − 1)3 gives n = 30. There are 30 two-digit numbers divisible by 3 — counted without listing one of them.

Example-8 handles counting from the far end. In 10, 7, 4, ..., −62 there are 25 terms, so the 11th term from the last is the 15th from the start, not the 14th. The book flags this trap itself.

5. Gauss, and the Sum of n Terms

Hema puts Rs 1000 into her daughter's money box on her first birthday and increases it by Rs 500 every year. How much is in the box by her 21st birthday? Adding twenty-one numbers is possible but, as the book says, tedious.

The chapter's answer is the trick attributed to Gauss, who as a boy was asked for the sum of the integers from 1 to 100 and replied 5050.

The same move works on any A.P. Write Sₙ forwards and backwards, add the two term by term, and every column becomes 2a + (n − 1)d. There are n columns, so 2Sₙ = n[2a + (n − 1)d] and

Sₙ = (n/2)[2a + (n − 1)d]

Since a + (n − 1)d is just the last term, the same formula can be written Sₙ = (n/2)(a + l) — the count times the average of the first and last terms. When the first and last terms are known but d is not, this second form is the one to reach for.

For Hema's money box, a = 1000, d = 500 and n = 21 give S = (21/2)[2000 + 10000] = Rs 1,26,000. Setting a = 1 and l = n gives the special case the book boxes: the sum of the first n positive integers is n(n + 1)/2.

The book also records a Remark worth keeping: since Sₙ includes everything up to the nth term and Sₙ₋₁ everything up to the one before, aₙ = Sₙ − Sₙ₋₁. Exercise 6.3 question 8 turns that into a whole problem, recovering an A.P. from the rule Sₙ = 4n − n².

6. Two Answers, and How to Read Them

Example-12 asks how many terms of 24, 21, 18, ... must be taken so that their sum is 78. The equation 78 = (n/2)[51 − 3n] reduces to n² − 17n + 52 = 0, which factorises as (n − 4)(n − 13) = 0.

Both n = 4 and n = 13 are admissible, and the book explains why rather than leaving it strange. The terms from the 5th to the 13th sum to zero: since a is positive and d negative, the later terms turn negative and cancel the positive ones exactly.

That is a different situation from the rejected roots of chapter 5. Here neither answer is impossible — both are genuinely correct, and the right response is to give both.

Exercise 6.3 question 13 is the contrast. Stacking 200 logs with 20 in the bottom row and one fewer in each row above gives n² − 41n + 400 = 0 and the roots 16 and 25. But 25 rows would make the top row 20 − 24 = −4 logs. Here one root must be discarded, and the working is the same as before; only the situation decides.

7. Geometric Progressions

Section 6.5 changes the operation. In 30, 90, 270, 810 each term is the preceding one multiplied by 3; in 1/4, 1/16, 1/64, 1/256 the multiplier is 1/4; in 30, 24, 19.2, 15.36, 12.288 it is 0.8.

Such a list is a geometric progression, and the fixed multiplier is the common ratio r. The general form is a, ar, ar², ar³, ... and the test for a G.P. is that consecutive ratios agree: a₂/a₁ = a₃/a₂ = ... = r.

The book attaches conditions to r that are worth noticing. It requires a ≠ 0, r ≠ 0 and r ≠ 1, and repeats all three in the chapter summary.

The last of those is a convention choice. Many other books allow r = 1 and call the resulting constant list a G.P., so a student cross-reading two books will meet a disagreement here. Note also the contrast with A.P.s, where this same book is happy to call 2, 2, 2, 2 an arithmetic progression with d = 0.

Geometric progressions turn up wherever something grows or shrinks by a proportion. A chain letter sent to four friends gives 1, 4, 16, 64, 256. Rs 500 at 10% compounded annually gives 550, 605, 665.5. A pendulum whose arc is 0.9 of the previous swing gives 18, 16.2, 14.58, 13.122.

The nth term follows the same way as for an A.P., by counting the operations. The second term is ar, the third ar², the fourth ar³, so

aₙ = a rⁿ⁻¹

and again the exponent is n − 1, not n, because the first term is multiplied by nothing. Example-20 applies it to 5/2, 5/4, 5/8, where r = 1/2, giving a₂₀ = 5/2²⁰ and the general term 5/2ⁿ.

Example-21 runs it backwards. Which term of 2, 2√2, 4, ... is 128? With r = √2 the equation 2(√2)ⁿ⁻¹ = 128 becomes 2^((n−1)/2) = 2⁶, so n − 1 = 12 and 128 is the 13th term.

Example-22 recovers both parameters from two terms. If a₃ = 24 and a₆ = 192, dividing one by the other gives r³ = 8, so r = 2 and a = 6, and then a₁₀ = 6 × 2⁹ = 3072. Dividing to eliminate a is the geometric counterpart of subtracting to eliminate a in an A.P.

8. The Formulas Side by Side

Notice what the table does not contain. The chapter derives the sum of an A.P. but gives no formula for the sum of a geometric progression. Geometric sums are left to a later class, so every G.P. question here is about terms and ratios, never about totals.

9. Where the Progressions Show Up

Exercise 6.1 sorts situations rather than lists. A taxi charging Rs 20 for the first kilometre and Rs 8 for each additional one gives 20, 28, 36 — an A.P. A vacuum pump removing 1/4 of the remaining air each time leaves 3/4 of the air each time, so the amounts form a G.P., not an A.P. Money at compound interest is likewise geometric.

Exercise 6.3 gathers the sum applications. Seven cash prizes totalling Rs 700, each Rs 20 less than the one before, work out to 160, 140, 120, 100, 80, 60 and 40. Three sections of each class from I to XII planting trees numbered as their class give 3 × 78 = 234 trees.

Two are worth doing carefully. A spiral of thirteen semicircles with radii 0.5, 1.0, ..., 6.5 cm has total length π times the sum of the radii, and since a semicircle of radius r has length πr, that is π × 45.5 = 143 cm with π taken as 22/7.

Exercise 6.2 collects the term questions, and several are worth naming because they recur in examinations. Finding which term of 3, 8, 13, ... equals 78 gives the 16th. Counting the three-digit multiples of 7 gives 128, and the multiples of 4 between 10 and 250 give 60. Asking when the nth terms of 63, 65, 67, ... and 3, 10, 17, ... coincide gives n = 13.

Two of them are about the difference rather than the terms. If the 17th term exceeds the 10th by 7, then 7d = 7 and d = 1. And if two A.P.s share a common difference and their 100th terms differ by 100, then their 1000th terms differ by 100 as well, because the shared d cancels and only the gap between the first terms survives.

The Optional Exercise pushes further. A ladder with rungs 25 cm apart, shortening from 45 cm to 25 cm over a span of 2.5 m, has 250/25 + 1 = 11 rungs, so the wood needed is (11/2)(45 + 25) = 385 cm. And houses numbered 1 to 49 have exactly one house, number 35, for which the numbers before it sum to the same as the numbers after it.

10. What the Exercises Ask, and Where the Book Goes Wrong

Five exercises carry this chapter. Exercise 6.1 has four questions on recognising A.P.s, Exercise 6.2 has seventeen on the nth term, Exercise 6.3 has fourteen on sums, Exercise 6.4 has four on recognising G.P.s and Exercise 6.5 has seven on the nth term of a G.P., with seven more in the Optional Exercise.

The printed answers run from textbook page 378 to page 381. Exercises 6.1, 6.3 and 6.5 are entirely correct, which is the best run in the book so far, and Exercise 6.2 is right except for one answer's form. But one printed answer is simply wrong.

Exercise 6.4 question 4 asks for x so that x, x + 2, x + 6 are consecutive terms of a geometric progression, and the key answers −4. Substituting −4 gives the list −4, −2, 2, whose ratios are 1/2 and −1. Those differ, so it is not a G.P. at all.

The condition (x + 2)² = x(x + 6) expands to x² + 4x + 4 = x² + 6x, so 4 = 2x and x = 2, giving 2, 4, 8 with ratio 2 throughout.

Exercise 6.2 question 17 asks in which year Subba Rao's salary reached Rs 7000, given that he started in 1995 at Rs 5000 with a Rs 200 annual increment. The key answers 11. That is the term number, not a year: the eleventh year of his service is 2005. The arithmetic is right and the question is not answered.

One more is worth flagging as ambiguous rather than wrong. Exercise 6.4 question 1(i) asks whether Sharmila's salary forms a G.P., given Rs 5,00,000 in the first year and a "yearly increment of 10%", and the key answers No.

That is defensible only if the increment is a fixed 10% of the first year's salary, which would make it an arithmetic progression. Read the usual way, as 10% of the current salary, the salaries are 500000, 550000, 605000 — a G.P. with r = 1.1. The question needed the word "compounded", which the book supplies elsewhere when it means it.

11. Four Things in the Chapter Text

Two are misspelt names. The History box on page 131 credits the transmission of these progressions to early Greek writers on the authority of "Boethins", which appears to be a misprint for Boethius. Page 145 captions the portrait "Carl Fredrich Gauss", where the usual spelling is Friedrich.

One is a loose end. Among the opening patterns, item (ii) on page 129 asks for the next two terms of 1, 2, 4, 8, 10, 20, 22, ... — a list whose rule the chapter never states, before or after. It is not an A.P., not a G.P., and not addressed again.

The last is phrasing. Page 153 says a pendulum's arc becomes "0.9th part of the previous length", where "0.9 times" is meant. Smaller slips run through the chapter: "25sh year" on page 139, "4h year" on page 142, "resepectively" on page 153.

12. Summary

A pattern that repeats is not a progression. An arithmetic progression adds a fixed common difference d to each term; a geometric progression multiplies by a fixed common ratio r. Everything in this chapter follows from which of those two operations is at work.

To test a list, subtract consecutive terms for an A.P. and divide them for a G.P., always taking the later term against the earlier one. A single pair is not enough; check that the result is the same throughout.

The nth term of an A.P. is a + (n − 1)d, and of a G.P. is a rⁿ⁻¹. In both, the exponent or multiplier is n − 1 rather than n, because the first term has had the operation applied to it no times at all.

The sum of an A.P. comes from Gauss's pairing. Writing the sum forwards and backwards and adding gives Sₙ = (n/2)[2a + (n − 1)d], or equivalently (n/2)(a + l) when the last term is known. The book gives no sum formula for a G.P.

A quadratic in n can produce two admissible answers, as when 24, 21, 18, ... sums to 78 for both 4 terms and 13. It can also produce one that the situation forbids, as when 200 logs would need a row of −4. Read each root against the problem, and say which case you are in.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

General form of an A.P.

a, a + d, a + 2d, a + 3d, ...

Two numbers generate the whole list: the first term a and the common difference d.

Test for an A.P.

a2 - a1 = a3 - a2 = ... = a(k+1) - ak = d

Always subtract the earlier term from the later one. In 6, 3, 0, -3 the difference is 3 - 6 = -3, not 6 - 3.

nth term of an A.P.

aₙ = a + (n - 1)d

The exponent of the operation is n - 1, not n, because the first term has had d added to it zero times.

Last term of a finite A.P.

l = a + (m - 1)d, where m is the number of terms

Finding m from this is the first step in any question about a term counted from the end.

Sum of the first n terms of an A.P.

Sₙ = n/2 [2a + (n - 1)d]

Derived by writing the sum forwards and backwards and adding, so that all n columns become 2a + (n - 1)d.

Sum when the last term is known

Sₙ = n/2 (a + l)

The count times the average of the first and last terms. Use this whenever a and l are given but d is not.

Sum of the first n positive integers

1 + 2 + 3 + ... + n = n(n + 1)/2

The special case a = 1, l = n. Gauss's 1 to 100 gives 100 times 101 over 2 = 5050.

Recovering a term from the sums

aₙ = Sₙ - S(n-1)

The book states this as a Remark. Exercise 6.3 question 8 uses it to rebuild an A.P. from Sₙ = 4n - n squared.

General form of a G.P.

a, ar, ar^2, ar^3, ...

This book requires a not 0, r not 0 and r not 1. The last condition is a convention choice; many other books allow r = 1.

Test for a G.P.

a2/a1 = a3/a2 = ... = aₙ / a(n-1) = r

Divide rather than subtract, and again take the later term over the earlier one.

nth term of a G.P.

aₙ = a r^(n-1)

Same n - 1 as in the A.P., and for the same reason: the first term is multiplied by r zero times.

Finding r from two known terms

divide one term equation by the other to eliminate a

From a3 = 24 and a6 = 192, dividing gives r cubed = 8, so r = 2. This is the geometric counterpart of subtracting to eliminate a in an A.P.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using n instead of n - 1 in the nth term formula

✓ The first term has had the common difference added to it zero times, so the nth term needs n - 1 additions. The 15th salary is 8000 plus 500 taken 14 times, not 15.

WATCH OUT

✗ Subtracting in the wrong direction to find d

✓ Always take the later term minus the earlier one. The book flags this directly: in 6, 3, 0, -3 the common difference is 3 - 6 = -3. Reversing it turns a decreasing A.P. into an increasing one.

WATCH OUT

✗ Declaring a list an A.P. after checking only one pair of terms

✓ Check at least two differences. In 2, 5, 7, 10, 12, 15 the first gap is 3, which looks promising, but the second is 2 and the list is not an A.P.

WATCH OUT

✗ Treating compound interest or a percentage increase as an arithmetic progression

✓ A fixed rupee increment is arithmetic; a fixed percentage is geometric, because the amount added grows with the balance. Exercise 6.1 question 1(iv) and the vacuum pump question both test this.

WATCH OUT

✗ Concluding a number belongs to an A.P. without checking that n is a whole number

✓ Solving 301 = 5 + (n - 1)6 gives n = 151/3. A fractional or negative n means the number is not a term of the list at all, and saying so is the complete answer.

WATCH OUT

✗ Counting a term from the end by subtracting instead of using the total

✓ In a 25-term A.P. the 11th term from the last is the 15th from the start, not the 14th. The rule is position from start = total - position from end + 1.

WATCH OUT

✗ Rejecting one root of a quadratic in n out of habit

✓ Sometimes both are right. For 24, 21, 18, ... summing to 78, both n = 4 and n = 13 hold, because the terms from the 5th to the 13th cancel out. Check what the situation actually forbids before discarding.

WATCH OUT

✗ Keeping a root that makes a count negative

✓ For 200 logs with 20 in the bottom row, n = 25 gives a top row of -4 logs and must go, leaving n = 16. Test each root against the physical meaning.

WATCH OUT

✗ Looking for a sum formula for a geometric progression in this chapter

✓ There is none. The chapter derives the sum of an A.P. only, and every G.P. question here is about terms and ratios. Geometric sums come in a later class.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Progressions?

23 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

23 questions~16 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A repeating pattern is not a progression; a progressive pattern changes by a fixed rule each step
•An A.P. adds a fixed common difference d; a G.P. multiplies by a fixed common ratio r
•The general form of an A.P. is a, a + d, a + 2d, a + 3d, ...
•To test for an A.P. subtract consecutive terms, always later minus earlier
•In 6, 3, 0, -3 the common difference is 3 - 6 = -3, not 6 - 3
•d may be positive, negative or zero, so 2, 2, 2, 2 is a valid A.P.
•An A.P. with a last term is finite; one that never ends is infinite
•The Fibonacci-style list 1, 1, 2, 3, 5 is NOT an A.P. - its gaps are 0, 1, 1, 2
•aₙ = a + (n - 1)d, with n - 1 because the first term has had d added zero times
•A fractional n means the number tested is not a term of the A.P. at all
•Counting how many numbers in a range are divisible by k is an nth-term question
•The kth term from the end of an m-term A.P. is the (m - k + 1)th from the start
•Gauss's method: write the sum forwards and backwards, add, and every column is the same
•Sₙ = n/2 [2a + (n - 1)d] and equivalently Sₙ = n/2 (a + l)
•Use the second form whenever the first and last terms are known but d is not
•The sum of the first n positive integers is n(n + 1)/2, and 1 to 100 gives 5050
•aₙ = Sₙ - S(n-1), which recovers the terms when only the sum formula is given
•Hema's money box holds Rs 1,26,000 on the 21st birthday, from a = 1000, d = 500, n = 21
•A quadratic in n can give two admissible answers: 24, 21, 18, ... sums to 78 at both n = 4 and n = 13
•It can also give one impossible answer: 200 logs need 16 rows, since 25 rows would mean a top row of -4
•A G.P. has general form a, ar, ar^2, ar^3, ... and nth term a r^(n-1)
•This book requires r not equal to 1 for a G.P., which many other books do not
•Test a G.P. by dividing consecutive terms, not subtracting
•Compound interest, depreciation and repeated halving are geometric; fixed increments are arithmetic
•Joining the midpoints of a square halves its area, giving 256, 128, 64, 32 as a G.P.
•Divide two term equations to eliminate a and find r, as in a3 = 24 and a6 = 192 giving r = 2
•This chapter gives NO sum formula for a G.P.; geometric sums come in a later class
•The printed answer key is wrong for Exercise 6.4 question 4, where the answer is 2 and not -4

Telangana (TSBIE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: No marks distribution is printed in the textbook for this chapter or anywhere in the volume, so no total is claimed. The index allots 11 periods in January. The categories below are the book's own five numbered exercises plus its Optional Exercise and its in-chapter Do This, Try This, Think and Discuss and Activity boxes; the marks column indicates question size rather than official weightage. Answers for Exercises 6.1 to 6.5 are printed at textbook pages 378 to 381. Exercises 6.1, 6.3 and 6.5 are entirely correct - the best run in the book so far - and Exercise 6.2 is correct except for the form of one answer. ONE PRINTED ANSWER IS WRONG: Exercise 6.4 question 4 asks for x so that x, x + 2, x + 6 are consecutive terms of a G.P. and answers -4, but -4 gives the list -4, -2, 2 whose ratios are 1/2 and -1, so it is not a G.P. at all; the condition (x + 2) squared = x(x + 6) gives x = 2 and the list 2, 4, 8. A second answer is incomplete rather than wrong: Exercise 6.2 question 17 asks in which YEAR Subba Rao's salary reached Rs 7000 and answers 11, which is the term number - the year is 2005. A third, Exercise 6.4 question 1(i), answers No to whether a salary with a 'yearly increment of 10%' forms a G.P.; that holds only if the increment is a fixed 10% of the first year's salary, whereas the usual reading of a percentage increment compounds and does give a G.P. with r = 1.1.

Question type	Marks each	Typical count
Exercise 6.1	22	4
Exercise 6.2	44	17
Exercise 6.3	44	14
Exercise 6.4	16	4
Exercise 6.5	24	7
Optional Exercise	26	7
In-chapter boxes	14	11

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Salary scales and pension slabs

Salary scales and pension slabs, where a fixed annual increment makes the yearly figures an arithmetic progression

Loan repayment by equal instalments

Loan repayment by equal instalments, and the reducing-balance schedules that sit behind them

Compound interest

Compound interest, inflation and population growth, all geometric because the change is proportional

Depreciation of machinery and vehicles

Depreciation of machinery and vehicles, which the chapter's Optional Exercise works out over ten years

Seating tiers

Seating tiers, stepped stands and stacked stock, where each row differs from the last by a fixed count

Construction estimates such as the wood in a ladder or th…

Construction estimates such as the wood in a ladder or the bricks in a staircase, which are sums of arithmetic progressions

Radioactive decay and drug clearance

Radioactive decay and drug clearance, where a fixed fraction remains after each equal interval

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write down a, d and n before touching any formula; most lost marks come from a misidentified d

For a decreasing list, write d as a negative number rather than remembering to subtract later

Choose Sₙ = n/2 (a + l) whenever the last term is given, and the 2a + (n - 1)d form otherwise

When a question gives two terms, subtract the equations for an A.P. and divide them for a G.P.

For a term counted from the end, find the total number of terms first and write the position conversion explicitly

If solving for n gives a fraction, say plainly that the number is not a term - that is the full answer, not a dead end

When a quadratic in n gives two roots, state both and then say which the situation allows and why

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Prove that if the pth term of an A.P. is q and the qth term is p, then the (p + q)th term is zero

STRETCH

Show that the sum of the first n odd numbers is n squared, and find the corresponding result for the first n even numbers

STRETCH

If a, b, c are in A.P. and also in G.P., prove that a = b = c

STRETCH

Prove that three numbers are in G.P. exactly when the square of the middle one equals the product of the outer two, and use this to solve the x, x + 2, x + 6 problem in one line

STRETCH

Find all arithmetic progressions of three positive integers whose sum is 21 and whose product is a perfect cube

STRETCH

Investigate which totals can be made by stacking logs as in Exercise 6.3 question 13, and characterise when the resulting quadratic in n has a usable root

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

Telangana SSC public examination - Mathematics Paper I, where an nth-term question and a sum word problem on progressions are both standing items

Navodaya and Telangana residential school entrance tests, which favour pattern-continuation and simple nth-term questions

NTSE and state mathematics talent tests, where progressions appear as quick term-and-sum items and as number-pattern puzzles

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because nothing changes from one petal to the next. The book draws this line in its first paragraphs: natural patterns like petals, honeycomb cells and pine cone spirals have a repetition which is not progressive - the petals are equidistantly grown and the hexagons are identical. A progression needs each term to differ from the one before by a fixed rule, as Usha's salary does when Rs 500 is added every year. Repetition is sameness; progression is regular change.

Because the first term has had nothing done to it yet. To reach the second term you add d once, to reach the third you add it twice, and to reach the nth you add it n - 1 times. The book shows this concretely with Usha's salary: her 15th-year salary is 8000 plus 500 added fourteen times, which is 8000 + 14 times 500 = 15000. The same reasoning gives the n - 1 exponent in the geometric formula a r^(n-1).

Assume it is the nth term, solve for n, and look at what you get. Example-6 tests whether 301 belongs to 5, 11, 17, 23, ... by solving 301 = 5 + (n - 1)6, which gives n = 151/3. Since a term position must be a positive whole number, 301 is not in the list. A fractional n, or a negative one, is a complete and correct answer to this kind of question - you do not need to keep looking.

Sometimes both, and sometimes neither can be chosen by habit. In Example-12, 24 + 21 + 18 + ... equals 78 at both n = 4 and n = 13, and the book explains why: the terms from the 5th to the 13th add to zero, because a is positive and d is negative so the later terms turn negative and cancel the earlier ones. Both answers are genuinely correct. But in the 200-logs question, n = 25 would mean a top row of -4 logs, which is impossible, so only n = 16 survives. Test each root against the situation rather than against a rule.

No, it is geometric. A fixed rupee increment is arithmetic because the same amount is added each time; a fixed percentage is geometric because the amount added grows with the balance. Rs 500 at 10% compounded annually gives 550, 605, 665.5 - each term is the previous one times 1.1, and the differences 50, 55, 60.5 are not constant. Exercise 6.1 asks exactly this about Rs 10000 at 8%, and the answer is that it does not form an A.P.

It is not in this chapter. The book derives the sum of an arithmetic progression through Gauss's method and gives both Sₙ = n/2 [2a + (n - 1)d] and Sₙ = n/2 (a + l), but for geometric progressions it stops at the nth term. Its own summary lists only the general form and a r^(n-1). Every G.P. question in Exercises 6.4 and 6.5 is about terms, ratios or which term equals a given value; none asks for a total. Geometric sums are taken up in a later class.

Mostly, and more than for earlier chapters. Exercises 6.1, 6.3 and 6.5 are entirely correct, which no previous chapter managed. But one answer is wrong: Exercise 6.4 question 4 gives -4 for the x that makes x, x + 2, x + 6 a G.P., and -4 produces -4, -2, 2 with ratios 1/2 and -1, which is not a G.P. The answer is 2, giving 2, 4, 8. Also, Exercise 6.2 question 17 answers 11 to a question asking for a year; 11 is the term number and the year is 2005.

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Related chapters

Students who read Progressions usually read these next.

Real Numbers

The division algorithm, Euclid's algorithm for the HCF, the Fundamental Theorem of Arithmetic, and then the two things those theorems let you prove — exactly which fractions have decimals that stop, and why the square root of 2 cannot be written as a fraction at all. Closes with exponents and the laws of logarithms. Telangana SCERT Class 10 Mathematics Chapter 1.

Sets

What makes a collection a set, the two ways of writing one down, and then the whole vocabulary — empty set, universal set, subset, union, intersection, difference, equal sets, finite and infinite, and cardinality — with Venn diagrams doing the explaining throughout. Telangana SCERT Class 10 Mathematics Chapter 2.

Polynomials

What a polynomial is and what its degree means, what a zero is and where it sits on the graph, the relationships that tie zeroes to coefficients for quadratics and cubics, and the division algorithm that lets one known zero unlock the rest. Telangana SCERT Class 10 Mathematics Chapter 3.

Pair of Linear Equations in Two Variables

Why one equation in two unknowns is never enough, what the three pictures of a pair of lines mean, how the ratios of the coefficients predict the answer before you draw anything, and the substitution and elimination methods that finish the job when the graph will not. Telangana SCERT Class 10 Mathematics Chapter 4.

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