Exercise Set 8.1
1. Find the first five terms of the sequence in which the n-th term is given by:
(i) tn = 3n – 4
• t1 = 3(1) – 4 = -1
• t2 = 3(2) – 4 = 2
• t3 = 3(3) – 4 = 5
• t4 = 3(4) – 4 = 8
• t5 = 3(5) – 4 = 11
First five terms: -1, 2, 5, 8, 11
(ii) tn = 2 – 5n
• t1 = 2 – 5(1) = -3
• t2 = 2 – 5(2) = -8
• t3 = 2 – 5(3) = -13
• t4 = 2 – 5(4) = -18
• t5 = 2 – 5(5) = -23
First five terms: -3, -8, -13, -18, -23
(iii) tn = n2 – 2n + 3
• t1 = (1)2 – 2(1) + 3 = 2
• t2 = (2)2 – 2(2) + 3 = 3
• t3 = (3)2 – 2(3) + 3 = 6
• t4 = (4)2 – 2(4) + 3 = 11
• t5 = (5)2 – 2(5) + 3 = 18
First five terms: 2, 3, 6, 11, 18
2. Find the 10th and 15th terms of the sequence tn = 5n – 3 for n ≥ 1.
• t10 = 5(10) – 3 = 50 – 3 = 47
• t15 = 5(15) – 3 = 75 – 3 = 72
10th term = 47, 15th term = 72
3. Determine whether 97 and 172 are terms of the sequence tn = 5n – 3 for n ≥ 1.
For 97:
5n – 3 = 97 ⟹ 5n = 100 ⟹ n = 20
Since 20 is a positive integer, 97 is the 20th term.
For 172:
5n – 3 = 172 ⟹ 5n = 175 ⟹ n = 35
Since 35 is a positive integer, 172 is the 35th term.
Yes, both 97 and 172 are terms of the sequence (20th and 35th terms respectively).
4. Which term of the sequence tn = 5n – 3 for n ≥ 1 is 607?
5n – 3 = 607
5n = 610
n = 122
607 is the 122nd term of the sequence.
5. A sequence is given by the recursive rule t1 = -5, tn+1 = tn + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
• t1 = -5
• t2 = t1 + 3 = -5 + 3 = -2
• t3 = t2 + 3 = -2 + 3 = 1
• t4 = t3 + 3 = 1 + 3 = 4
• t5 = t4 + 3 = 4 + 3 = 7
Explicit formula: tn = -5 + (n – 1)3 = 3n – 8
To check if 52 is a term: 3n – 8 = 52 ⟹ 3n = 60 ⟹ n = 20
First five terms: -5, -2, 1, 4, 7. Yes, 52 is the 20th term of the sequence.
6. Let T1 = 1, T2 = 2, T3 = 4, and Tn = Tn-1 + Tn-2 + Tn-3 for n ≥ 4. Find T4, T5, T6, T7, and T8.
• T4 = T3 + T2 + T1 = 4 + 2 + 1 = 7
• T5 = T4 + T3 + T2 = 7 + 4 + 2 = 13
• T6 = T5 + T4 + T3 = 13 + 7 + 4 = 24
• T7 = T6 + T5 + T4 = 24 + 13 + 7 = 44
• T8 = T7 + T6 + T5 = 44 + 24 + 13 = 81
T4 = 7, T5 = 13, T6 = 24, T7 = 44, T8 = 81
Exercise Set 8.2
1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
First term a = 3, common difference d = 8 – 3 = 5
• a10 = a + 9d = 3 + 9(5) = 3 + 45 = 48
• a26 = a + 25d = 3 + 25(5) = 3 + 125 = 128
10th term = 48, 26th term = 128
2. Which term of the AP: 21, 18, 15, … is -81? Also, is 0 a term of this AP? Give reasons for your answer.
a = 21, d = 18 – 21 = -3
For -81:
an = a + (n – 1)d ⟹ -81 = 21 + (n – 1)(-3)
-102 = (n – 1)(-3) ⟹ n – 1 = 34 ⟹ n = 35
For 0:
0 = 21 + (n – 1)(-3) ⟹ -21 = (n – 1)(-3) ⟹ n – 1 = 7 ⟹ n = 8
-81 is the 35th term. 0 is the 8th term of the AP because 8 is a positive integer.
3. Find the n-th term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
a = 11, d = 8 – 11 = -3
• Explicit n-th term formula: an = 11 + (n – 1)(-3) = 14 – 3n
• Recursive rule: t1 = 11, tn+1 = tn – 3 for n ≥ 1
n-th term = 14 – 3n; Recursive rule: t1 = 11, tn+1 = tn – 3
4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
a + 2d = 12 — (Equation 1)
a + 49d = 106 — (Equation 2)
Subtracting Equation 1 from Equation 2: 47d = 94 ⟹ d = 2
Substituting d = 2 into Equation 1: a + 2(2) = 12 ⟹ a = 8
29th term: a29 = a + 28d = 8 + 28(2) = 8 + 56 = 64
29th term = 64
5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
2-digit numbers divisible by 3: 12, 15, 18, …, 99
a = 12, d = 3, last term l = 99
99 = 12 + (n – 1)3 ⟹ 87 = (n – 1)3 ⟹ n – 1 = 29 ⟹ n = 30
Sum Sn = (n / 2)(a + l) = (30 / 2)(12 + 99) = 15 × 111 = 1665
There are 30 two-digit numbers divisible by 3. Their sum is 1665.
6. Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?
a = 500000, d = 20000, an = 700000
700000 = 500000 + (n – 1)20000
200000 = (n – 1)20000 ⟹ n – 1 = 10 ⟹ n = 11
n = 11 corresponds to the 11th year (after 10 completed years of increments).
His income reached ₹7,00,000 in the 11th year (after 10 years).
7. A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?
a = 1, d = 1, n = 25
S25 = (n(n + 1)) / 2 = (25 × 26) / 2 = 25 × 13 = 325
The child uses 325 marbles in all.
Exercise Set 8.3
1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
r = 2, a8 = ar7 = 192 ⟹ a(2)7 = 192 ⟹ 128a = 192 ⟹ a = 1.5
a12 = ar11 = a8 × r4 = 192 × 24 = 192 × 16 = 3072
12th term = 3072
2. Find the 10th and n-th terms of the GP: 5, 25, 125, ….
a = 5, r = 25 / 5 = 5
• an = arn-1 = 5 × 5n-1 = 5n
• a10 = 510 = 9765625
n-th term = 5n, 10th term = 510 = 9,765,625
3. A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn – 2 for n ≥ 1. Which term of the sequence is 730?
• t1 = 2
• t2 = 3(2) – 2 = 4
• t3 = 3(4) – 2 = 10
• t4 = 3(10) – 2 = 28
• t5 = 3(28) – 2 = 82
Pattern analysis: tn = 3n-1 + 1
To find n for 730: 3n-1 + 1 = 730 ⟹ 3n-1 = 729 = 36 ⟹ n – 1 = 6 ⟹ n = 7
730 is the 7th term of the sequence.
4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the n-th term.
a = 2, r = 3
• Explicit formula: an = 2 × 3n-1
• Recursive formula: t1 = 2, tn+1 = 3tn for n ≥ 1
To find n: 2 × 3n-1 = 4374 ⟹ 3n-1 = 2187 = 37 ⟹ n – 1 = 7 ⟹ n = 8
4374 is the 8th term. Explicit: an = 2 × 3n-1. Recursive: t1 = 2, tn+1 = 3tn.
5. A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way—each time rising to 60% of the previous height.
(i) What height does the ball reach after the 5th bounce?
Height after n-th bounce hn = 80 × (0.6)n
h5 = 80 × (0.6)5 = 80 × 0.07776 = 6.2208 metres
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Total distance = Initial drop + 2 × (Sum of heights of first 5 bounces)
Sum of 5 bounces S5 = 80(0.6)[1 – (0.6)5] / (1 – 0.6) = 48 × (1 – 0.07776) / 0.4 = 120 × 0.92224 = 110.6688 m
Total distance = 80 + 2(110.6688) = 80 + 221.3376 = 301.3376 metres
(i) 6.2208 metres; (ii) 301.3376 metres
6. Which term of the sequence 2, 2√2, 4, … is 128?
a = 2, r = 2√2 / 2 = √2 = 21/2
an = a × rn-1 = 2 × (21/2)n-1 = 21 + (n-1)/2 = 128 = 27
1 + (n – 1) / 2 = 7 ⟹ (n – 1) / 2 = 6 ⟹ n – 1 = 12 ⟹ n = 13
128 is the 13th term.
7. Sierpiński Square Carpet Questions
(i) How many red squares are there in Stages 0 to 3?
Stage 0: 1, Stage 1: 8, Stage 2: 64, Stage 3: 512
(ii) Can you predict the number of red squares in Stages 4 and 5?
Stage 4: 84 = 4,096
Stage 5: 85 = 32,768
(iii) Can you find a rule for the number of red squares at the n-th stage? Write explicit and recursive formulas.
Explicit formula: Nn = 8n
Recursive formula: N0 = 1, Nn = 8 × Nn-1 for n ≥ 1
(iv) Area of red region analysis (Initial Area = 1 unit2):
Area at Stage n = (8/9)n
• Stage 1: 8/9, Stage 2: 64/81, Stage 3: 512/729
• Stage 4: (8/9)4 = 4096 / 6561 ≈ 0.6243
• Stage 5: (8/9)5 = 32768 / 59049 ≈ 0.5549
• Explicit formula: An = (8/9)n
• Recursive formula: A0 = 1, An = (8/9) × An-1
• As n increases, the area approaches 0.
(i) 1, 8, 64, 512; (ii) 4096, 32768; (iii) Nn = 8n; (iv) An = (8/9)n, approaches 0.
End-of-Chapter Exercises
1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
a + 10d = 38 — (Eq 1)
a + 15d = 73 — (Eq 2)
Subtracting Eq 1 from Eq 2: 5d = 35 ⟹ d = 7
Substituting d = 7 into Eq 1: a + 70 = 38 ⟹ a = -32
a31 = a + 30d = -32 + 30(7) = -32 + 210 = 178
31st term = 178
2. Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
a + 2d = 16 — (Eq 1)
a7 – a5 = 12 ⟹ (a + 6d) – (a + 4d) = 12 ⟹ 2d = 12 ⟹ d = 6
From Eq 1: a + 2(6) = 16 ⟹ a = 4
The AP is 4, 10, 16, 22, 28, …
3. How many three-digit numbers are divisible by 7?
Smallest 3-digit number divisible by 7 = 105; Largest = 994
a = 105, d = 7, l = 994
994 = 105 + (n – 1)7 ⟹ 889 = 7(n – 1) ⟹ n – 1 = 127 ⟹ n = 128
There are 128 three-digit numbers divisible by 7.
4. How many multiples of 4 lie between 10 and 250?
First multiple after 10 = 12; Last multiple before 250 = 248
a = 12, d = 4, l = 248
248 = 12 + (n – 1)4 ⟹ 236 = 4(n – 1) ⟹ n – 1 = 59 ⟹ n = 60
There are 60 multiples of 4 between 10 and 250.
5. Find a GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.
a + ar = -4 ⟹ a(1 + r) = -4 — (Eq 1)
ar4 = 4(ar2) ⟹ r2 = 4 ⟹ r = 2 or r = -2
• Case 1: r = 2 ⟹ a(3) = -4 ⟹ a = -4/3. GP: -4/3, -8/3, -16/3, …
• Case 2: r = -2 ⟹ a(1 – 2) = -4 ⟹ -a = -4 ⟹ a = 4. GP: 4, -8, 16, -32, …
GPs: (-4/3, -8/3, -16/3, …) OR (4, -8, 16, -32, …)
6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Sum of k consecutive terms starting at a: k[2a + (k – 1)] / 2 = 100 ⟹ k(2a + k – 1) = 200
k must be a factor of 200 such that k < 2a + k – 1 and both give integer ‘a’ ≥ 1.
• k = 5: 5(2a + 4) = 200 ⟹ 2a + 4 = 40 ⟹ 2a = 36 ⟹ a = 18. (18+19+20+21+22 = 100)
• k = 8: 8(2a + 7) = 200 ⟹ 2a + 7 = 25 ⟹ 2a = 18 ⟹ a = 9. (9+10+11+12+13+14+15+16 = 100)
2 ways: (18+19+20+21+22) and (9+10+11+12+13+14+15+16)
7. Bacteria culture doubles every hour (initial = 30). How many at end of 2nd, 4th, and n-th hour?
Count at end of t hours N(t) = 30 × 2t
• End of 2nd hour: 30 × 22 = 120
• End of 4th hour: 30 × 24 = 480
• End of n-th hour: 30 × 2n
2nd hour = 120, 4th hour = 480, n-th hour = 30 × 2n
8. Sum of 4th and 8th terms of AP is 24; sum of 6th and 10th terms is 44. Find first three terms.
(a + 3d) + (a + 7d) = 24 ⟹ 2a + 10d = 24 ⟹ a + 5d = 12 — (Eq 1)
(a + 5d) + (a + 9d) = 44 ⟹ 2a + 14d = 44 ⟹ a + 7d = 22 — (Eq 2)
Subtracting Eq 1 from Eq 2: 2d = 10 ⟹ d = 5
a + 5(5) = 12 ⟹ a = -13
First three terms: -13, -8, -3
9. Smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
n(n + 1) / 2 > 1000 ⟹ n(n + 1) > 2000
Testing n = 44: 44 × 45 = 1980 (< 2000)
Testing n = 45: 45 × 46 = 2070 (> 2000)
Smallest value of n is 45.
10. Which term of GP: 2, 8, 32, … is 131072? Write explicit and recursive formulas.
a = 2, r = 4
• Explicit formula: an = 2 × 4n-1
• Recursive formula: t1 = 2, tn+1 = 4tn
2 × 4n-1 = 131072 ⟹ 4n-1 = 65536 = 48 ⟹ n – 1 = 8 ⟹ n = 9
131072 is the 9th term. Explicit: an = 2 × 4n-1; Recursive: t1 = 2, tn+1 = 4tn
11. Sum of first three terms of GP is 13/12 and their product is -1. Find common ratio and terms.
Let terms be a/r, a, ar. Product = a3 = -1 ⟹ a = -1.
Sum: -1/r – 1 – r = 13/12 ⟹ -(1/r + r) = 25/12 ⟹ (r2 + 1)/r = -25/12
12r2 + 25r + 12 = 0 ⟹ (3r + 4)(4r + 3) = 0 ⟹ r = -4/3 or r = -3/4
• If r = -4/3: Terms are 3/4, -1, 4/3
• If r = -3/4: Terms are 4/3, -1, 3/4
Common ratio r = -4/3 or -3/4. Terms are 3/4, -1, 4/3 (or in reverse order).
12. If 4th, 10th and 16th terms of a GP are x, y, and z respectively, prove that x, y, z are in GP.
x = ar3, y = ar9, z = ar15
y / x = (ar9) / (ar3) = r6
z / y = (ar15) / (ar9) = r6
Since y / x = z / y = r6, x, y, z are in Geometric Progression.
Proved: y / x = z / y = r6, hence x, y, z form a GP.
13. Sum of first three terms of GP is 26, and sum of their squares is 364. Find terms.
a + ar + ar2 = 26 ⟹ a(1 + r + r2) = 26 — (Eq 1)
a2 + a2r2 + a2r4 = 364 ⟹ a2(1 + r2 + r4) = 364 — (Eq 2)
Note that 1 + r2 + r4 = (1 + r + r2)(1 – r + r2)
Divide Eq 2 by (Eq 1)2: (1 – r + r2) / (1 + r + r2) = 364 / 676 = 7 / 13
13(1 – r + r2) = 7(1 + r + r2) ⟹ 6r2 – 20r + 6 = 0 ⟹ 3r2 – 10r + 3 = 0
(3r – 1)(r – 3) = 0 ⟹ r = 3 or r = 1/3
• For r = 3: a(1 + 3 + 9) = 26 ⟹ 13a = 26 ⟹ a = 2. Terms: 2, 6, 18
The terms of the GP are 2, 6, 18 (or 18, 6, 2).
14. P1 = 1, P2 = 2 and for n > 2, Pn = P1 + P2 + … + Pn-1 + 1. Find the values of P1, P2, …, P8. Can you find a simpler recursive formula for Pn? Can you give an explicit formula?
• P1 = 1
• P2 = 2
• P3 = (1 + 2) + 1 = 4
• P4 = (1 + 2 + 4) + 1 = 8
• P5 = 16, P6 = 32, P7 = 64, P8 = 128
Observations: Pn = 2Pn-1 for n ≥ 3 with P2 = 2
• Simpler recursive formula: P1 = 1, P2 = 2, Pn = 2Pn-1 for n ≥ 3
• Explicit formula: P1 = 1, and Pn = 2n-1 for n ≥ 2
Values: 1, 2, 4, 8, 16, 32, 64, 128. Explicit: P1 = 1, Pn = 2n-1 (for n ≥ 2).
15. W1 = 1, W2 = 2 and for n > 2, Wn = W1 + W2 + … + Wn-2 + 2. Find the values of W1, W2, …, W8. Do you recognise this sequence?
• W1 = 1
• W2 = 2
• W3 = W1 + 2 = 1 + 2 = 3
• W4 = W1 + W2 + 2 = 1 + 2 + 2 = 5
• W5 = (1 + 2 + 3) + 2 = 8
• W6 = (1 + 2 + 3 + 5) + 2 = 13
• W7 = (1 + 2 + 3 + 5 + 8) + 2 = 21
• W8 = (1 + 2 + 3 + 5 + 8 + 13) + 2 = 34
Sequence generated: 1, 2, 3, 5, 8, 13, 21, 34, …
Values: 1, 2, 3, 5, 8, 13, 21, 34. This is the Fibonacci Sequence (starting from the 2nd Fibonacci number).
