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State Board · Class 10 · Mathematics Part 1

Lesson 3: Arithmetic Progression

5 concepts · 40 practice questions with answers · Free to read

This lesson is also in: मराठी

Sequence

From daily life

Seats in a bus are numbered 1, 2, 3, ... and each passenger has a fixed place. Students standing in a line for assembly also have a first place, a second place and so on. Numbers arranged in a fixed order like this form a sequence.

Simple definition

A sequence (श्रेढी) is a list of numbers in a definite order. Its terms are t1, t2, t3, ..., tn, ... where tn is the term at the nth place.

Rule 1: Terms and notation — The sequence is written {tn}. If it is infinite, there is a term tn for every positive integer n. In 3, 6, 9, 12, ... we have t1 = 3, t2 = 6, t4 = 12. In 1, 4, 9, 16, 25, ... the term at place n is n2, so t6 = 36.

  • Easy : Easy: In 5, 10, 15, 20, ... t3 = 15.
  • Medium : Medium: In 1, 8, 27, 64, ... (cubes) the 5th term is 53 = 125.
  • Exam-type : Exam: If tn = 3n + 1, then t1 = 4, t2 = 7, t3 = 10 and t10 = 31.

Rule 2: Sequences with a rule — The list 4, 17, −2, 0, 9, ... is a sequence too, but its terms follow no rule. In general we study only sequences with a rule that decides the next term, such as 4, 8, 12, 16, ... or 2, 4, 8, 16, ... or 12, 13, 14, 15, ...

  • Easy : Easy: In 2, 4, 6, 8, ... the rule is "add 2".
  • Medium : Medium: In 3, 6, 12, 24, ... the rule is "multiply by 2".
  • Exam-type : Exam: Next two terms of 12, 13, 14, ... are 15 and 16.
Remember

tn is the term at place n. The order matters.

Common mistake

Writing tn as t × n. It is a name for the nth term.

Shortcut

Memory clue: "स्थान n = पद tn".

Spot the difference

SequenceRulet5
3, 6, 9, ...Add 315
2, 4, 8, ...Multiply by 232
1, 4, 9, ...n225

Practice questions

Easy
  1. 1What is a sequence?
  2. 2What does tn denote?
  3. 3In 7, 14, 21, 28, ... find t2.
Medium
  1. 4Next term of 5, 10, 15, 20?
  2. 5Next term of 2, 4, 8, 16?
  3. 6tn = 2n − 1. Find t5.
Challenging
  1. 7tn = n2 + 1. Write the first three terms.
  2. 8Why do we study only sequences with a rule?
Show answers (try them yourself first)
  1. 1Numbers arranged in a definite order.
  2. 2The term at the nth place.
  3. 314.
  4. 425.
  5. 532.
  6. 69.
  7. 72, 5, 10.
  8. 8Only then can the next terms be predicted.

Arithmetic Progression

From daily life

On a cycle ride you cover 5 km on day 1, 8 km on day 2, 11 km on day 3, adding 3 km each day. A steady "add the same number" is the heart of an A.P.

Simple definition

If tn+1 − tn is the same for every n, the sequence is an Arithmetic Progression (A.P.; समांतर श्रेढी) and that constant is the common difference (सामाईक फरक) d.

Rule 1: Common difference — d can be positive, negative or zero. 7, 10, 13, 16, ... has d = 3. 50, 45, 40, 35, ... has d = −5. 6, 6, 6, 6, ... has d = 0. To find d, subtract any term from the next.

An arithmetic progression 3, 7, 11, 15, 19 where 4 is added at every step3a+ 47a + d+ 411a + 2d+ 415a + 3d+ 419a + 4dA.P.: the same number is added every time (d = 4)

In the figure : In the figure an A.P. adds the same number d at every step (here d = 4).

  • Easy : Easy: 4, 9, 14, 19: d = 5, next terms 24, 29.
  • Medium : Medium: 20, 17, 14, 11: d = −3, next terms 8, 5.
  • Exam-type : Exam: 12, 1, 32, 2, ... is an A.P. with d = 12; next terms 52 and 3.

Rule 2: General form — An A.P. with first term a and common difference d is a, (a + d), (a + 2d), (a + 3d), ... Here t1 = a, t2 = a + d, t3 = a + 2d. For a = 5, d = −2 the A.P. is 5, 3, 1, −1, ... For a = −4, d = 6 it is −4, 2, 8, 14, ...

  • Easy : Easy: a = 2, d = 3 gives 2, 5, 8, 11.
  • Medium : Medium: a = 10, d = −4 gives 10, 6, 2, −2.
  • Exam-type : Exam: Is 3, 7, 12, 18 an A.P.? Differences 4, 5, 6 are not equal, so no.

Rule 3: A.P. or geometric progression? — Adding the same number gives an A.P. Multiplying by the same number gives a geometric progression (G.P.; गुणोत्तर श्रेढी). Pocket money ₹ 10, 20, 30, 40 is an A.P.; ₹ 10, 20, 40, 80 doubles each time, so it is a G.P. The squares 1, 4, 9, 16 are neither. An A.P. can be finite (₹ 100, 200, ..., 1200 saved in a year) or infinite (multiples of 5: 5, 10, 15, ...).

An A.P. adds the same number, a G.P. multiplies by the same numberAdding or multiplying?A.P. (add)2, 5, 8, 11, ...add 3 each timed = 3G.P. (multiply)2, 6, 18, 54, ...multiply by 3 each timenot an A.P.

In the figure : In the figure adding the same number gives an A.P.; multiplying by the same number gives a geometric progression.

  • Easy : Easy: 2, 4, 6, 8 is an A.P.
  • Medium : Medium: 3, 9, 27, 81 is a G.P. (×3).
  • Exam-type : Exam: 1, 1, 2, 2, 3, 3 is not an A.P., because the differences are 0, 1, 0, 1, 0.
Remember

A.P.: same difference each step. G.P.: same multiplier each step.

Common mistake

Subtracting the next term from the previous one. Use tn+1 − tn (next minus previous).

Shortcut

Memory clue: "मिळवा = A.P.; गुणा = G.P.".

Spot the difference

TypeRuleExample
A.P.Add d7, 10, 13
G.P.Multiply by r3, 9, 27
NeitherNo constant rule1, 4, 9

Practice questions

Easy
  1. 1What is an A.P.?
  2. 2Find d for 11, 15, 19, 23.
  3. 3Find d for 30, 26, 22, ...
Medium
  1. 4Write the A.P. for a = 6, d = 5 (4 terms).
  2. 5Is 2, 6, 18, 54 an A.P.?
  3. 6Next two terms of 9, 5, 1, ...
Challenging
  1. 7Show that 1, 3, 6, 10 is not an A.P.
  2. 8Is 32, 12, −12, −32 an A.P.? Find d.
Show answers (try them yourself first)
  1. 1A sequence in which the difference of consecutive terms is constant.
  2. 24.
  3. 3−4.
  4. 46, 11, 16, 21.
  5. 5No, it is a G.P. (×3).
  6. 6−3, −7.
  7. 7Differences 2, 3, 4 are not equal.
  8. 8Yes, d = −1.

nth term of an A.P.

From daily life

To reach step 20 of a staircase that rises by 3 cm at each step, you do not climb all 20 one by one in your head. You start at the ground level and add the step height 19 times. The nth term formula does exactly this.

Simple definition

In an A.P. with first term a and common difference d, the nth term is tn = a + (n − 1)d.

Rule 1: The formula — t2 = a + d = a + (2 − 1)d, t3 = a + 2d = a + (3 − 1)d, and so tn = a + (n − 1)d. From a we add d exactly (n − 1) times. For 4, 7, 10, 13, ... a = 4 and d = 3, so t20 = 4 + 19 × 3 = 61.

Bars 4, 7, 10, 13, 16, 19: every bar adds one more d to the first term a4t17t210t313t416t519t6Each step adds one more d: tₙ = a + (n − 1)dadt₆ = a + 5d = 4 + 5 × 3 = 19

In the figure : In the figure the nth term: starting from a, we add d exactly (n − 1) times.

  • Easy : Easy: 2, 5, 8, ...: t10 = 2 + 9 × 3 = 29.
  • Medium : Medium: 15, 13, 11, ...: t12 = 15 + 11 × (−2) = −7.
  • Exam-type : Exam: 3, 8, 13, ...: tn = 3 + (n − 1) × 5 = 5n − 2, so t30 = 148.

Rule 2: Is a number in the A.P.? How many terms? — Put tn equal to the number and solve for n; n must be a natural number. In 3, 8, 13, ...: is 98 a term? 3 + (n − 1) × 5 = 98 gives n = 20, so yes, the 20th term. Is 100? (n − 1) × 5 = 97 gives n − 1 = 19.4, so no. Number of terms in 7, 12, 17, ..., 102: 102 = 7 + (n − 1) × 5, n = 20.

  • Easy : Easy: Terms in 3, 6, 9, ..., 45: 45 = 3 + (n − 1) × 3, n = 15.
  • Medium : Medium: Which term of 5, 11, 17, ... is 83? 83 = 5 + (n − 1) × 6, n = 14.
  • Exam-type : Exam: Two-digit multiples of 7 are 14, 21, ..., 98: 98 = 14 + (n − 1) × 7, n = 13.

Rule 3: Finding a and d from two terms — If t5 = 19 and t9 = 35, then a + 4d = 19 and a + 8d = 35. Subtract: 4d = 16, d = 4, and a = 19 − 16 = 3. So tn = 3 + (n − 1) × 4 = 4n − 1 and t20 = 79.

  • Easy : Easy: a = 10, t2 = 14 gives d = 4.
  • Medium : Medium: t3 = 11 and t7 = 27 give 4d = 16, d = 4, a = 3.
  • Exam-type : Exam: t6 = 30 and t10 = 10 give 4d = −20, d = −5, a = 55.
Remember

tn = a + (n − 1)d. Add d only (n − 1) times, not n.

Common mistake

Using a + nd. That gives tn+1.

Shortcut

Memory clue: "पहिले + (n − 1) × फरक".

Spot the difference

ItemFormula
nth terma + (n − 1)d
Number of termsn = l − ad + 1
Two terms givenSubtract to get d

Practice questions

Easy
  1. 1Write the formula for tn.
  2. 26, 9, 12, ...: find t8.
  3. 340, 35, 30, ...: find t9.
Medium
  1. 4How many terms in 5, 10, 15, ..., 100?
  2. 5Is 50 a term of 4, 9, 14, ...?
  3. 6t4 = 14 and t8 = 26. Find a and d.
Challenging
  1. 7How many two-digit numbers are divisible by 6?
  2. 8Which term of 2, 9, 16, ... is 261?
Show answers (try them yourself first)
  1. 1a + (n − 1)d.
  2. 227.
  3. 30.
  4. 420.
  5. 54 + (n − 1) × 5 = 50 gives n − 1 = 9.2, so no.
  6. 6d = 3, a = 5.
  7. 712, 18, ..., 96: n = 15.
  8. 8261 = 2 + (n − 1) × 7, n = 38.

Sum of n terms of an A.P.

From daily life

Young Gauss was asked to add 1 to 100. He paired 1 + 100, 2 + 99, 3 + 98, ...: 50 pairs, each 101, so 50 × 101 = 5050. Pairing the first and last term is the whole secret of A.P. sums.

Simple definition

The sum of the first n terms of an A.P. is Sn = n2 [2a + (n − 1)d] = n2 (first term + last term).

Rule 1: Gauss's idea and the formula — Write Sn forwards and backwards and add: 2Sn = n[2a + (n − 1)d]. So Sn = n2 [2a + (n − 1)d]. Since the last term is tn = a + (n − 1)d, also Sn = n2 (a + tn). Johann Carl Friedrich Gauss (1777 − 1855) was a great German mathematician.

Numbers 1 to 10 paired as 1 with 10, 2 with 9 and so on; each pair adds to 11 and the total is 55Pair the first with the last: every pair adds up to 111101129113811471156115 pairs × 11 = 55 = 1 + 2 + 3 + ... + 10

In the figure : In the figure Gauss's pairing idea shown with 1 to 10: 5 pairs, each adding to 11.

  • Easy : Easy: 1 + 2 + ... + 20 = 20 × 212 = 210.
  • Medium : Medium: 5, 9, 13, ...: S10 = 5 × (10 + 36) = 230.
  • Exam-type : Exam: 3 + 7 + 11 + ... + 43: n = 11 and S11 = 112 × (3 + 43) = 253.

Rule 2: Special sums and shortcut problems — First n natural numbers: n(n + 1)2. First n even numbers: n(n + 1); for 25 even numbers, 25 × 26 = 650. First n odd numbers: n2; for 15 odd numbers, 225. Two-digit multiples of 5: 10, 15, ..., 95; n = 18 and S = 9 × 105 = 945. Three terms in A.P. with sum 21 and product 315: take a − d, a, a + d; 3a = 21, a = 7; 7(49 − d2) = 315 gives d2 = 4, so the terms are 5, 7, 9.

Live figure

Sum of n terms of an A.P.

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Open in the app
  • Easy : Easy: First 10 odd numbers add up to 100.
  • Medium : Medium: a = 4, d = 3, n = 12: S = 6 × (8 + 33) = 246.
  • Exam-type : Exam: 5 + 7 + 9 = 21 and 5 × 7 × 9 = 315, which verifies the three-term answer.
Remember

Sn = n ÷ 2 × (first + last). First find n if it is not given.

Common mistake

Multiplying n by (first + last) without halving. Do not forget n2.

Shortcut

Memory clue: "जोड्यांची संख्या × जोडीची बेरीज".

Spot the difference

SumResult
1 + 2 + ... + nn(n + 1)2
First n even numbersn(n + 1)
First n odd numbersn2

Practice questions

Easy
  1. 1Who found 1 + 2 + ... + 100 by pairing?
  2. 2Write Sn using a and d.
  3. 3Sum of 1 to 50?
Medium
  1. 4Sum of first 12 odd numbers?
  2. 5Sum of first 10 even numbers?
  3. 6a = 2, d = 4, n = 8. Find S8.
Challenging
  1. 7Find 6 + 10 + 14 + ... + 50.
  2. 8Three terms in A.P. have sum 24 and product 480. Find them.
Show answers (try them yourself first)
  1. 1Gauss.
  2. 2n2 [2a + (n − 1)d].
  3. 31275.
  4. 4144.
  5. 5110.
  6. 64 × (4 + 28) = 128.
  7. 7n = 12, S = 6 × 56 = 336.
  8. 8a = 8, 8(64 − d2) = 480, d2 = 4; terms 6, 8, 10.

Applications of A.P.

From daily life

A tailor stitches 5 shirts on day 1 and 2 more every next day. To know how many shirts are ready in two weeks, he uses A.P. formulas instead of counting day by day.

Simple definition

Many real situations with a steady increase or decrease are A.P.s. Find a and d, then use tn and Sn.

Rule 1: Production and saving — A factory makes 400 chairs in year 2 and 550 in year 5, with the same yearly increase. Then a + d = 400 and a + 4d = 550 give d = 50 and a = 350. In year 10: 350 + 9 × 50 = 800. In 5 years: S5 = 52 × (700 + 200) = 2250 chairs.

Five monthly savings 150, 200, 250, 300, 350 in an A.P. with total 1250150t1200t2250t3300t4350t5Savings of ₹ 150, 200, 250, ... in 5 monthsadS₅ = 5/2 × (150 + 350) = ₹ 1250

In the figure : In the figure application: monthly savings in an A.P. and their total.

  • Easy : Easy: Saving ₹ 100, ₹ 120, ₹ 140, ...: a = 100, d = 20.
  • Medium : Medium: The 12th month saving is 100 + 11 × 20 = ₹ 320.
  • Exam-type : Exam: In 12 months the total is 122 × (100 + 320) = ₹ 2520.

Rule 2: Repaying a loan — A person repays ₹ 6000 by paying ₹ 1100 in month 1 and ₹ 100 less each month. Sn = n2 [2200 − 100(n − 1)] = 6000 gives n2 − 23n + 120 = 0, that is (n − 8)(n − 15) = 0. For n = 15 the payments would turn negative (t15 = −300), so n = 8. Check: t8 = 400 and S8 = 4 × 1500 = 6000.

  • Easy : Easy: First payment 1100, second 1000: d = −100.
  • Medium : Medium: t8 = 1100 + 7 × (−100) = 400.
  • Exam-type : Exam: Reject n = 15, because a payment cannot be negative.

Rule 3: Rows and shapes — An auditorium has 20 rows; the first row has 15 seats and each next row has 2 more. t20 = 15 + 19 × 2 = 53 and S20 = 10 × (15 + 53) = 680 seats. Trees planted 1, 2, 3, ... in 20 rows total 20 × 212 = 210.

  • Easy : Easy: Row 2 has 17 seats.
  • Medium : Medium: Row 10 has 15 + 9 × 2 = 33 seats.
  • Exam-type : Exam: 10 rows hold 102 × (15 + 33) = 240 seats.
Remember

Find a and d first. Reject answers that do not make sense (negative n or negative amounts).

Common mistake

Giving n = 15 as an answer when payments become negative. Check each root.

Shortcut

Memory clue: "a, d शोधा → tn किंवा Sn वापरा → अर्थ तपासा".

Spot the difference

Situationad
Chair production35050
Monthly saving10020
Loan repayment1100−100

Practice questions

Easy
  1. 1How do we spot an A.P. in a story?
  2. 2A saver puts ₹ 50, ₹ 70, ₹ 90, ... What are a and d?
  3. 3How much in the 10th month?
Medium
  1. 4Total in 10 months?
  2. 5A theatre has 10 rows with 12, 14, 16, ... seats. Seats in row 10?
  3. 6Total seats there?
Challenging
  1. 7t3 = 600, t7 = 800 for yearly production. Find a and d.
  2. 8Why do we reject a root that gives negative payments?
Show answers (try them yourself first)
  1. 1A steady, equal increase or decrease each time.
  2. 2a = 50, d = 20.
  3. 350 + 9 × 20 = ₹ 230.
  4. 45 × (50 + 230) = ₹ 1400.
  5. 530.
  6. 65 × (12 + 30) = 210.
  7. 7d = 50, a = 500.
  8. 8A payment cannot be negative, so that value does not fit the story.

Quick revision

  • A.P.: tn+1 − tn = d is constant and tn = a + (n − 1)d.
  • Sn = n2 [2a + (n − 1)d] = n2 (first term + last term).
  • The sum of the first n natural numbers is n(n + 1)2, of the first n even numbers is n(n + 1) and of the first n odd numbers is n².