Class 9 · MathsNCERT 2026 Revised Syllabus
Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions (0904CH08)
Ganita Manjari Part I · 35 questions · 4 exercises
Chapter code: 0904ch08 · Book: Ganita Manjari Part I · Class 9 Maths
Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part I Chapter 8. Every exercise has a clear, step-by-step answer for chapter code 0904CH08.
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Exercise Set 8.1(6 questions)
1Find the first five terms of the sequence in which the nth term is given by(i) tn = 3n – 4,(ii) tn = 2 – 5n, and(iii) tn = n2 – 2n + 3 for n ≥ 1.
School Level Solution
i) tₙ=3n−4: −1, 2, 5, 8, 11.ii) tₙ=2−5n: −3, −8, −13, −18, −23.iii) tₙ=n²−2n+3: 2, 3, 4, 7, 12.
2Find the 10th and 15th terms of the sequence tn = 5n – 3 for n ≥ 1.
School Level Solution
tₙ=5n−3t₁₀=5(10)−3=47.t₁₅=5(15)−3=72.
3Determine whether 97 and 172 are terms of the sequence tn = 5n – 3 for n ≥ 1.
School Level Solution
For tₙ=5n−3:97=5n−3 ⇒ 5n=100 ⇒ n=20. Hence 97 is a term.172=5n−3 ⇒ 5n=175 ⇒ n=35. Hence 172 is a term.
4Which term of the sequence tn = 5n – 3 for n ≥ 1 is 607?
School Level Solution
607=5n−3⇒ 5n=610⇒ n=122 Ans.
5A 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?
School Level Solution
t₁=−5 and tₙ₊₁=tₙ+3.First five terms = −5, −2, 1, 4, 7.For 52 to be a term: −5+3(n−1)=52⇒ 3(n−1)=57 ⇒ n−1=19 ⇒ n=20.
6Let T1 = 1, T2 = 2, T3 = 4, and Tn = Tn–1 + Tn–2 + Tn–3 for n ≥ 4. Find T4, T5, T6, T7, and T8 .
School Level Solution
T₁=1, T₂=2, T₃=4T₄=T₃+T₂+T₁=4+2+1=7.T₅=7+4+2=13.T₆=13+7+4=24.T₇=24+13+7=44.T₈=44+24+13=81.
Exercise Set 8.2(7 questions)
1Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
School Level Solution
AP: 3,8,13,18,…a=3, d=5.tₙ=a+(n−1)dt₁₀=3+9×5=48.t₂₆=3+25×5=128.
2Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer.
School Level Solution
a=21, d=−3.−81=21+(n−1)(−3)−102=−3(n−1)n−1=34 ⇒ n=35.For 0: 0=21−3(n−1) gives n=8, so 0 is a term.
3Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
School Level Solution
a=11, d=−3.tₙ=11+(n−1)(−3)=14−3n.∴ explicit rule = tₙ=14−3n.Recursive rule: t₁=11, tₙ₊₁=tₙ−3.
4An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term. (Hint: If ‘a’ is the first term and ‘d’ the common difference, then we arrive at the equations a + 2d = 12 and a + 49d = 106. Solve this pair of linear equations for ‘a’ and ‘d’.)
School Level Solution
a+2d=12 ...(1)a+49d=106 ...(2)Subtracting: 47d=94 ⇒ d=2.a+4=12 ⇒ a=8.t₂₉=8+28×2=64 Ans.
5How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
School Level Solution
Smallest 2-digit multiple of 3 = 12; largest = 99.AP: 12,15,…,99.99=12+(n−1)3 ⇒ n=30.Sum = n2(a+l) = 302(12+99) = 1665.
6Harish 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?
School Level Solution
Initial salary = ₹5,00,000; increment = ₹20,000 per year.5,00,000 + 20,000n = 7,00,000⇒ 20,000n=2,00,000⇒ n=10 years.
7A 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?
School Level Solution
Number of marbles = 1+2+3+…+25.S=252(1+25)=325 marbles.
Exercise Set 8.3(7 questions)
1
Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
School Level Solution
Given r=2 and 8th term = 192.t₈=ar⁷192=a×2⁷a=192128=32.t₁₂=(32)×2¹¹=3072 Ans.
2Find the 10th and nth terms of the GP: 5, 25, 125, … .
School Level Solution
GP: 5,25,125,…a=5, r=5.t₁₀=5×5⁹=5¹⁰.tₙ=5×5ⁿ⁻¹=5ⁿ.
★3A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn – 2 for n ≥ 1. Which term of the sequence is 730?
School Level Solution
Given, t₁=2 and tₙ₊₁=3tₙ−2.
Now,
t₂=3(2)−2=4
t₃=3(4)−2=10
t₄=3(10)−2=28
t₅=3(28)−2=82
t₆=3(82)−2=244
t₇=3(244)−2=730
Therefore, 730 is the 7th term.
4Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.
School Level Solution
GP: 2,6,18,…a=2, r=3.2×3ⁿ⁻¹=43743ⁿ⁻¹=2187=3⁷∴ n=8.Explicit formula: tₙ=2×3ⁿ⁻¹.Recursive formula: t₁=2, tₙ₊₁=3tₙ.
5A 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?(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
School Level Solution
Initial height = 80 m and the bounce ratio is r=0.6.
(i) Height after the 5th bounce = 80(0.6)⁵ = 6.2208 m.
(ii) The first hit on the ground occurs after the initial 80 m drop. To reach the 6th hit, the ball then travels up and down through the first five bounce heights: 48, 28.8, 17.28, 10.368 and 6.2208 m.
Total distance = 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208) = 301.3376 m.
6Which term of the sequence 2, 2 2 , 4 , … is 128?
School Level Solution
Sequence: 2, 2√2, 4,…a=2, r=√2.2(√2)ⁿ⁻¹=128(√2)ⁿ⁻¹=64=2⁶⇒ (n−1)/2=6∴ n=13.
7Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on. Stage 0 Stage 1 Stage 2 Stage 3 Fig. 8.12: Stages 0, 1, 2 and 3 of the Sierpiński square carpet Look at Fig. 8.12 and try to answer the following questions.(i) How many red squares are there in Stages 0 to 3?(ii) Can you predict the number of red squares in Stages 4 and 5?(iii) Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.(iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?
School Level Solution
Red squares at stages 0,1,2,3 = 1, 8, 64, 512.ii) Stage 4 = 8⁴ = 4096, Stage 5 = 8⁵ = 32768.iii) Number of red squares at nth stage = 8ⁿ; recursive rule: R₀=1, Rₙ=8Rₙ₋₁.iv) Area of red region = (89)ⁿ square units.Stages 1,2,3: 89, 6481, 512729.Stages 4,5: 40966561 and 3276859049.As n increases, the area tends to 0.
End-of-Chapter Exercises(15 questions)
1Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
School Level Solution
t₁₁=38 ⇒ a+10d=38 ...(1)t₁₆=73 ⇒ a+15d=73 ...(2)Subtracting: 5d=35 ⇒ d=7.a+70=38 ⇒ a=−32.t₃₁=−32+30×7=178.
2Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
School Level Solution
t₃=a+2d=16t₇−t₅=2d=12 ⇒ d=6.a+12=16 ⇒ a=4.∴ AP = 4,10,16,22,….
★3How many three-digit numbers are divisible by 7? (Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)
School Level Solution
Smallest 3-digit no. divisible by 7 = 105
Largest 3-digit no. divisible by 7 = 994
These numbers form an AP.
a=105, l=994, d=7
l=a+(n−1)d
994=105+(n−1)7
889=7(n−1)
n−1=127
n=128
There are 128 three-digit numbers divisible by 7.
★4How many multiples of 4 lie between 10 and 250? (Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.)
School Level Solution
First multiple of 4 after 10 = 12
Last multiple of 4 before 250 = 248
a=12, l=248, d=4
l=a+(n−1)d
248=12+(n−1)4
236=4(n−1)
n−1=59
n=60
There are 60 multiples of 4 between 10 and 250.
★5Find a GP for which the sum of the first two terms is – 4 and the fifth term is 4 times the third term.
School Level Solution
Let the GP be a, ar, ar², …
Given, a+ar=−4
a(1+r)=−4 ...(1)
Also, fifth term = 4×third term.
ar⁴=4ar²
r²=4
⇒ r=2 or −2
When r=2, from (1), 3a=−4 ⇒ a=−43
GP = −43, −83, −163, −323, −643
When r=−2, from (1), −a=−4 ⇒ a=4
GP = 4, −8, 16, −32, 64
★6Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
School Level Solution
Let the consecutive natural nos. be
a, a+1, a+2, … a+(n−1)
Now, ATQ,
Sₙ = 100
Sₙ = a + a+1 + a+2 + …. + a+(n−1)
100 = a + a+1 + a+2 + …. + a+(n−1)
100 = n2(a+a+(n−1)) [∵ Sₙ = n2(a+l)]
200 = n(2a+n−1)
There can be following two cases:
(i) When n is odd, 2a+n−1 = even
(ii) When n is even, 2a+n−1 = odd
We see that, in both cases the combination becomes odd × even.
200 = odd × even
Now, 200 = 2×2×2×5×5
Factorising it into odd × even combinations, we get:
200 = 8×25, 200 = 5×40, 200 = 1×200
Now, let n=8 (from the above factor),
2a+n−1=25
2a+8−1=25
2a+7=25
2a=18
⇒ a=9
One of the possible ways of expressing 100 as a sum of consecutive natural nos.
=9+10+11+12+13+14+15+16
=100
Now, Putting n=25, we get
2a+(25−1)=8
2a+24=8
2a=−16
⇒ a=−8
Since the question asks natural no. and here, (a=−8) ∉ N, therefore we would not consider this sequence.
We could now put the smaller side value of n, only as larger one would always give negative.
Let n=5 (from the factor 5×40)
2a+5−1=40
2a=36
⇒ a=18
Another possible way is
18+19+20+21+22
=100
Putting a=1 does not becomes a series. Therefore we neglect it.
There are above two ways of expressing 100 as a sum of consecutive natural nos.
★7The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?
School Level Solution
Initial no. of bacteria in the tube = 30
per hr.
GP becomes:
30, 60, 120, …
Here, a=30
r=2
Here, n will be (hours+1), because 2nd term is at the end of 1st hour while 3rd term is at the end of 2nd hour.
OR
We can put directly the value of hours but use the formula [a rⁿ] instead of arⁿ⁻¹.
Now,
Bacteria at the end of 2nd hour = arⁿ
=30×2²
=30×4
=120 Ans.
Bacteria at the end of 4th hour = arⁿ
=30×2⁴
=30×16
=480 Ans.
Bacteria at the end of nth hour = arⁿ
=30×2ⁿ
=15×2ⁿ⁺¹ Ans.
★8The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
School Level Solution
Let first term=a and common difference=d.
t₄=a+3d, t₈=a+7d
t₄+t₈=24
2a+10d=24 ...(1)
t₆+t₁₀=44
2a+14d=44 ...(2)
Subtracting (1) from (2):
4d=20
d=5
2a+50=24
2a=−26
a=−13
First three terms are −13, −8, −3.
★9Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
School Level Solution
Sum of first n natural nos. = n(n+1)2
ATQ,
n(n+1)2 > 1000
n(n+1) > 2000
For n=44, 44×452=990 < 1000
For n=45, 45×462=1035 > 1000
∴ Smallest value of n = 45.
★10Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.
School Level Solution
GP: 2, 8, 32, …
a=2, r=4
tₙ=arⁿ⁻¹
2×4ⁿ⁻¹=131072
4ⁿ⁻¹=65536=4⁸
∴ n=9
Explicit formula: tₙ=2×4ⁿ⁻¹
Recursive formula: t₁=2, tₙ₊₁=4tₙ
★11The sum of the first three terms of a GP is 1312 and their product is −1. Find the common ratio and the terms.
School Level Solution
Let the three terms of the GP be a, ar, ar².Given, sum=1312 and product=−1.ar×ar×ar=−1⇒ ar=−1∴ the middle term = −1.Now, a−1+1a=1312a+1a=251212a²−25a+12=0(3a−4)(4a−3)=0a=43 or 34∴ The possible GPs are 43,−1,−34 and 34,−1,−43.
★12If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
School Level Solution
Let the GP be a, ar, ar², …
x=t₄=ar³
y=t₁₀=ar⁹
z=t₁₆=ar¹⁵
Now,
y²=(ar⁹)²=a²r¹⁸
xz=(ar³)(ar¹⁵)=a²r¹⁸
∴ y²=xz, hence x, y, z are in GP.
★13The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.
School Level Solution
Let the GP be a, ar, ar².
Given, a+ar+ar²=26 ...(1)
a²+a²r²+a²r⁴=364
=a²(1+r²+r⁴)=364 ...(2)
Solving (1) and (2), we get the three terms
2, 6, 18 (in one order).
Check: 2+6+18=26 and 4+36+324=364.
★14Suppose 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?
School Level Solution
P₁=1, P₂=2
P₃=P₁+P₂+1=1+2+1=4
P₄=1+2+4+1=8
P₅=1+2+4+8+1=16
Therefore,
P₁,P₂,…,P₈ = 1,2,4,8,16,32,64,128
Simpler recursive formula: Pₙ=2Pₙ₋₁, n>2.
Explicit formula: Pₙ=2ⁿ⁻¹.
★15Suppose 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?
School Level Solution
W₁=1, W₂=2
For n>2, Wₙ=W₁+W₂+…+Wₙ₋₂+2.
W₃=W₁+2=1+2=3
W₄=W₁+W₂+2=1+2+2=5
W₅=W₁+W₂+W₃+2=1+2+3+2=8
Similarly,
1,2,3,5,8,13,21,34
Do you recognise this sequence?
It is the Fibonacci sequence.