Exercise 5.1 answers: Arithmetic Progressions

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(i) ₹15 for the first km and ₹8 for each additional km

1. Fare for 1 km = ₹15.
2. Fare for 2 km = ₹15 + ₹8 = ₹23.
3. Fare for 3 km = ₹23 + ₹8 = ₹31.
4. Each extra km adds the same ₹8, so this is an AP with a = 15, d = 8.

AnswerYes, it is an AP (a = 15, d = 8).

(ii) removes 1/4 of the air remaining in the cylinder at a time

1. Say the cylinder starts with 1 unit of air.
2. Each time, 1/4 of what is left is removed, so 3/4 is left.
3. After 1 pump: 3/4. After 2 pumps: 3/4 × 3/4 = 9/16. After 3 pumps: 27/64.
4. The amount is multiplied by 3/4 each time instead of being reduced by a fixed amount, so the gaps −1/4, −3/16, −9/64 keep shrinking.

AnswerNo, it is not an AP — the air is multiplied by a fixed fraction each time, not reduced by a fixed amount.

(iii) ₹150 for the first metre … rises by ₹50 for each subsequent metre

1. Cost after 1 m = ₹150.
2. Cost after 2 m = ₹150 + ₹50 = ₹200.
3. Cost after 3 m = ₹200 + ₹50 = ₹250.
4. Each extra metre adds the same ₹50, so this is an AP with a = 150, d = 50.

AnswerYes, it is an AP (a = 150, d = 50).

(iv) ₹10000 is deposited at compound interest at 8 % per annum

1. Say ₹10000 is deposited.
2. After 1 year at 8% compound interest: ₹10000 × 1.08 = ₹10800.
3. After 2 years: ₹10800 × 1.08 = ₹11664. After 3 years: ₹11664 × 1.08 = ₹12597.12.
4. The gains ₹800, ₹864, ₹933.12 are not equal, because compound interest multiplies the growing amount by 1.08 each year instead of adding a fixed sum.

AnswerNo, it is not an AP — compound interest multiplies the amount by a fixed factor, not a fixed addend.

Watch this explained “Situations rather than lists”, 10:19 into Testing a given list, and what "finite" versus "infinite" changes

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(i) a = 10, d = 10

1. First term = 10.
2. Add d = 10 each time: 10, 20, 30, 40.

Answer10, 20, 30, 40

(ii) a = −2, d = 0

1. First term = −2.
2. Add d = 0 each time, so the list never moves: −2, −2, −2, −2.

Answer−2, −2, −2, −2

(iii) a = 4, d = −3

1. First term = 4.
2. Add d = −3 each time: 4, 1, −2, −5.

Answer4, 1, −2, −5

(iv) a = −1, d = 1/2

1. First term = −1.
2. Add d = 1/2 each time: −1, −1/2, 0, 1/2.

Answer−1, −1/2, 0, 1/2

(v) a = −1.25, d = −0.25

1. First term = −1.25.
2. Add d = −0.25 each time: −1.25, −1.5, −1.75, −2.

Answer−1.25, −1.5, −1.75, −2

Watch this explained “Two numbers rebuild everything”, 6:22 into A fixed step between neighbours is the whole definition

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(i) 3, 1, −1, −3, …

1. a = 3.
2. d = 1 − 3 = −2. Check: −1 − 1 = −2, −3 − (−1) = −2.

Answera = 3, d = −2

(ii) −5, −1, 3, 7, …

1. a = −5.
2. d = −1 − (−5) = 4. Check: 3 − (−1) = 4, 7 − 3 = 4.

Answera = −5, d = 4

(iii) 1/3, 5/3, 9/3, 13/3, …

1. a = 1/3.
2. d = 5/3 − 1/3 = 4/3. Check: 9/3 − 5/3 = 4/3, 13/3 − 9/3 = 4/3.

Answera = 1/3, d = 4/3

(iv) 0.6, 1.7, 2.8, 3.9, …

1. a = 0.6.
2. d = 1.7 − 0.6 = 1.1. Check: 2.8 − 1.7 = 1.1, 3.9 − 2.8 = 1.1.

Answera = 0.6, d = 1.1

Watch this explained “Later minus earlier”, 3:09 into A fixed step between neighbours is the whole definition

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(i) 2, 4, 8, 16, …

1. Differences: 4−2=2, 8−4=4, 16−8=8.
2. The differences 2, 4, 8 are not equal.

AnswerNo, not an AP.

(ii) 2, 5/2, 3, 7/2, …

1. Differences: 5/2−2=1/2, 3−5/2=1/2, 7/2−3=1/2.
2. All differences equal 1/2, so d = 1/2.
3. Next three terms: 7/2+1/2=4, 4+1/2=9/2, 9/2+1/2=5.

AnswerYes, AP with d = 1/2; next three terms 4, 9/2, 5.

(iii) −1.2, −3.2, −5.2, −7.2, …

1. Differences: −3.2−(−1.2)=−2, −5.2−(−3.2)=−2, −7.2−(−5.2)=−2.
2. All differences equal −2, so d = −2.
3. Next three terms: −7.2−2=−9.2, −9.2−2=−11.2, −11.2−2=−13.2.

AnswerYes, AP with d = −2; next three terms −9.2, −11.2, −13.2.

(iv) −10, −6, −2, 2, …

1. Differences: −6−(−10)=4, −2−(−6)=4, 2−(−2)=4.
2. All differences equal 4, so d = 4.
3. Next three terms: 2+4=6, 6+4=10, 10+4=14.

AnswerYes, AP with d = 4; next three terms 6, 10, 14.

(v) 3, 3+√2, 3+2√2, 3+3√2, …

1. Differences: (3+√2)−3=√2, (3+2√2)−(3+√2)=√2, (3+3√2)−(3+2√2)=√2.
2. All differences equal √2, so d = √2.
3. Next three terms: 3+4√2, 3+5√2, 3+6√2.

AnswerYes, AP with d = √2; next three terms 3+4√2, 3+5√2, 3+6√2.

(vi) 0.2, 0.22, 0.222, 0.2222, …

1. Differences: 0.22−0.2=0.02, 0.222−0.22=0.002, 0.2222−0.222=0.0002.
2. The differences 0.02, 0.002, 0.0002 are not equal.

AnswerNo, not an AP.

(vii) 0, −4, −8, −12, …

1. Differences: −4−0=−4, −8−(−4)=−4, −12−(−8)=−4.
2. All differences equal −4, so d = −4.
3. Next three terms: −12−4=−16, −16−4=−20, −20−4=−24.

AnswerYes, AP with d = −4; next three terms −16, −20, −24.

(viii) −1/2, −1/2, −1/2, −1/2, …

1. Every term is −1/2, so each difference is 0.
2. d = 0.
3. Next three terms are also −1/2, −1/2, −1/2 (the list never moves).

AnswerYes, AP with d = 0; next three terms −1/2, −1/2, −1/2.

(ix) 1, 3, 9, 27, …

1. Differences: 3−1=2, 9−3=6, 27−9=18.
2. The differences 2, 6, 18 are not equal (each term is 3 times the one before it — a fixed factor, not a fixed addend).

AnswerNo, not an AP.

(x) a, 2a, 3a, 4a, …

1. Differences: 2a−a=a, 3a−2a=a, 4a−3a=a.
2. All differences equal a, so d = a.
3. Next three terms: 4a+a=5a, 5a+a=6a, 6a+a=7a.

AnswerYes, AP with d = a; next three terms 5a, 6a, 7a.

(xi) a, a², a³, a⁴, …

1. Differences: a²−a, a³−a², a⁴−a³.
2. These are a(a−1), a²(a−1), a³(a−1), which are not equal to each other for a general value of a.

AnswerNo, not an AP (for a general value of a).

(xii) √2, √8, √18, √32, …

1. Simplify the surds first: √8=2√2, √18=3√2, √32=4√2.
2. So the list is √2, 2√2, 3√2, 4√2, and the differences are all √2.
3. d = √2. Next three terms: 5√2, 6√2, 7√2 (that is, √50, √72, √98).

AnswerYes, AP with d = √2; next three terms 5√2, 6√2, 7√2.

(xiii) √3, √6, √9, √12, …

1. Simplify the surds first: √9=3, √12=2√3; √6 does not simplify to a multiple of √3.
2. So the list is √3, √6, 3, 2√3, and the differences √6−√3, 3−√6, 2√3−3 are not equal.

AnswerNo, not an AP.

(xiv) 1², 3², 5², 7², …

1. The terms are 1, 9, 25, 49.
2. Differences: 9−1=8, 25−9=16, 49−25=24.
3. The differences 8, 16, 24 are not equal.

AnswerNo, not an AP.

(xv) 1², 5², 7², 73, …

1. The terms are 1, 25, 49, 73.
2. Differences: 25−1=24, 49−25=24, 73−49=24.
3. All differences equal 24, so d = 24.
4. Next three terms: 73+24=97, 97+24=121, 121+24=145.

AnswerYes, AP with d = 24; next three terms 97, 121, 145.

Watch this explained “The test itself: subtract along the list”, 0:36 into Testing a given list, and what "finite" versus "infinite" changes