Exercise Set 8.1 Solutions
Ganita Manjari | Grade 9 | Chapter 8: Sequences & Arithmetic Progressions
Q1. Find 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.
(i) tn = 3n − 4
For n = 1: t1 = 3(1) − 4 = 3 − 4 = −1
For n = 2: t2 = 3(2) − 4 = 6 − 4 = 2
For n = 3: t3 = 3(3) − 4 = 9 − 4 = 5
For n = 4: t4 = 3(4) − 4 = 12 − 4 = 8
For n = 5: t5 = 3(5) − 4 = 15 − 4 = 11
First 5 terms: −1, 2, 5, 8, 11
(ii) tn = 2 − 5n
For n = 1: t1 = 2 − 5(1) = 2 − 5 = −3
For n = 2: t2 = 2 − 5(2) = 2 − 10 = −8
For n = 3: t3 = 2 − 5(3) = 2 − 15 = −13
For n = 4: t4 = 2 − 5(4) = 2 − 20 = −17
For n = 5: t5 = 2 − 5(5) = 2 − 25 = −22
First 5 terms: −3, −8, −13, −17, −22
(iii) tn = n2 − 2n + 3
For n = 1: t1 = (1)2 − 2(1) + 3 = 1 − 2 + 3 = 2
For n = 2: t2 = (2)2 − 2(2) + 3 = 4 − 4 + 3 = 3
For n = 3: t3 = (3)2 − 2(3) + 3 = 9 − 6 + 3 = 6
For n = 4: t4 = (4)2 − 2(4) + 3 = 16 − 8 + 3 = 11
For n = 5: t5 = (5)2 − 2(5) + 3 = 25 − 10 + 3 = 18
First 5 terms: 2, 3, 6, 11, 18
Q2. Find the 10th and 15th terms of the sequence tn = 5n − 3 for n ≥ 1.
Given formula: tn = 5n − 3
To find 10th term (t10):
Substitute n = 10:
t10 = 5(10) − 3 = 50 − 3 = 47
To find 15th term (t15):
Substitute n = 15:
t15 = 5(15) − 3 = 75 − 3 = 72
Answer: 10th term = 47, 15th term = 72
Q3. Determine whether 97 and 172 are terms of the sequence tn = 5n − 3 for n ≥ 1.
Given: tn = 5n − 3
Part 1: Check for 97
Set tn = 97 ⇒ 5n − 3 = 97
5n = 97 + 3 ⇒ 5n = 100
n = 100 / 5 = 20 (Positive Integer)
✓ So, 97 is the 20th term of the sequence.
Part 2: Check for 172
Set tn = 172 ⇒ 5n − 3 = 172
5n = 172 + 3 ⇒ 5n = 175
n = 175 / 5 = 35 (Positive Integer)
✓ So, 172 is the 35th term of the sequence.
Answer: Both 97 and 172 are terms of the given sequence.
Q4. Which term of the sequence tn = 5n − 3 for n ≥ 1 is 607?
Given: tn = 607 and tn = 5n − 3
5n − 3 = 607
5n = 607 + 3
5n = 610
n = 610 / 5
n = 122
Answer: 607 is the 122nd term of the sequence.
Q5. 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?
Part 1: Finding the first 5 terms (Step-by-Step for each n)
Given recursive formula: tn+1 = tn + 3, with t1 = −5
• Given: t1 = −5
• For n = 1: Put n = 1 in formula ⇒ t1+1 = t1 + 3
⇒ t2 = −5 + 3 = −2
• For n = 2: Put n = 2 in formula ⇒ t2+1 = t2 + 3
⇒ t3 = −2 + 3 = 1
• For n = 3: Put n = 3 in formula ⇒ t3+1 = t3 + 3
⇒ t4 = 1 + 3 = 4
• For n = 4: Put n = 4 in formula ⇒ t4+1 = t4 + 3
⇒ t5 = 4 + 3 = 7
First 5 terms: −5, −2, 1, 4, 7
Part 2: Check if 52 is a term
Notice that the terms form an Arithmetic Progression (A.P.):
First term (a) = −5, Common difference (d) = 3
General term: tn = a + (n − 1)d
⇒ tn = −5 + (n − 1) × 3 = 3n − 8
Set tn = 52:
3n − 8 = 52 ⇒ 3n = 60 ⇒ n = 20
Answer: Yes, 52 is a term of this sequence, and it is the 20th term.
Q6. 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.
Given initial terms: T1 = 1, T2 = 2, T3 = 4
Recursive Rule: Tn = Tn−1 + Tn−2 + Tn−3
• For n = 4: Put n = 4 in formula
T4 = T3 + T2 + T1 = 4 + 2 + 1 = 7
• For n = 5: Put n = 5 in formula
T5 = T4 + T3 + T2 = 7 + 4 + 2 = 13
• For n = 6: Put n = 6 in formula
T6 = T5 + T4 + T3 = 13 + 7 + 4 = 24
• For n = 7: Put n = 7 in formula
T7 = T6 + T5 + T4 = 24 + 13 + 7 = 44
• For n = 8: Put n = 8 in formula
T8 = T7 + T6 + T5 = 44 + 24 + 13 = 81
Answer: T4 = 7, T5 = 13, T6 = 24, T7 = 44, T8 = 81
Q1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ....
Formula Used:
nth term of AP (a_n) = a + (n - 1)d
Where: a = First term, d = Common difference, n = Term number
Solution:
Given AP: 3, 8, 13, 18, ....
First term (a) = 3
Common difference (d) = 8 - 3 = 5
Finding the 10th term (n = 10):
a_10 = a + (10 - 1)d
a_10 = 3 + 9 × 5
a_10 = 3 + 45
a_10 = 48
Finding the 26th term (n = 26):
a_26 = a + (26 - 1)d
a_26 = 3 + 25 × 5
a_26 = 3 + 125
a_26 = 128
Answer: 10th term = 48, 26th term = 128
Q2. Which term of the AP: 21, 18, 15, ... is -81? Also, is 0 a term of this AP? Give reasons for your answer.
Concept:
Use a_n = a + (n - 1)d to find n. If n is a positive whole number, then the given value is a term of the AP.
Part 1: Is -81 a term?
Given AP: 21, 18, 15, ....
a = 21
d = 18 - 21 = -3
Let a_n = -81
a + (n - 1)d = -81
21 + (n - 1)(-3) = -81
(n - 1)(-3) = -81 - 21
(n - 1)(-3) = -102
n - 1 = -102 / -3
n - 1 = 34
n = 35
Since n is a positive integer, -81 is the 35th term.
Part 2: Is 0 a term?
Let a_n = 0
21 + (n - 1)(-3) = 0
(n - 1)(-3) = -21
n - 1 = -21 / -3
n - 1 = 7
n = 8
Since n = 8 is a positive integer, 0 is the 8th term of this AP.
Answer: -81 is the 35th term. Yes, 0 is the 8th term.
Q3. Find the nth term of the AP: 11, 8, 5, 2 ... Write the recursive rule for this AP.
Concept:
• General nth term: a_n = a + (n - 1)d
• Recursive rule: Expresses each term using the previous term (a_n = a_(n-1) + d)
Solution:
Given AP: 11, 8, 5, 2, ....
a = 11
d = 8 - 11 = -3
(i) Finding the nth term:
a_n = a + (n - 1)d
a_n = 11 + (n - 1)(-3)
a_n = 11 - 3n + 3
a_n = 14 - 3n
(ii) Recursive rule:
First term: a_1 = 11
Rule: a_n = a_(n-1) - 3 (for n ≥ 2)
Answer: nth term = 14 - 3n; Recursive Rule: a_1 = 11, a_n = a_(n-1) - 3
Q4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Total terms (n) = 50
3rd term (a_3) = 12 => a + 2d = 12 --- (Equation 1)
Last term (a_50) = 106 => a + 49d = 106 --- (Equation 2)
Subtracting Equation 1 from Equation 2:
(a + 49d) - (a + 2d) = 106 - 12
47d = 94
d = 94 / 47
d = 2
Substitute d = 2 in Equation 1:
a + 2(2) = 12
a + 4 = 12
a = 8
Finding the 29th term (a_29):
a_29 = a + 28d
a_29 = 8 + 28(2)
a_29 = 8 + 56
a_29 = 64
Answer: 29th term = 64
Q5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Formulas Used:
• Number of terms: a_n = a + (n - 1)d
• Sum of AP (S_n) = (n / 2) × (a + l) where l is the last term.
Solution:
2-digit numbers divisible by 3 are: 12, 15, 18, ..., 99
First term (a) = 12
Common difference (d) = 3
Last term (a_n or l) = 99
Step 1: Find total numbers (n)
a_n = a + (n - 1)d
99 = 12 + (n - 1)3
99 - 12 = 3(n - 1)
87 = 3(n - 1)
n - 1 = 87 / 3
n - 1 = 29
n = 30
Step 2: Find the sum (S_30)
S_30 = (30 / 2) × (12 + 99)
S_30 = 15 × 111
S_30 = 1665
Answer: Total numbers = 30, Sum = 1665
Q6. 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?
Solution:
Starting salary (a) = ₹5,00,000
Annual increment (d) = ₹20,000
Target salary (a_n) = ₹7,00,000
Finding n:
a_n = a + (n - 1)d
7,00,000 = 5,00,000 + (n - 1)20,000
7,00,000 - 5,00,000 = (n - 1)20,000
2,00,000 = (n - 1)20,000
n - 1 = 2,00,000 / 20,000
n - 1 = 10
n = 11
His salary reaches ₹7,00,000 in the 11th year (after 10 years of service).
Answer: After 10 years (in the 11th year)
Q7. 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?
Solution:
Number of marbles in rows form an AP: 1, 2, 3, 4, ..., 25
a = 1
d = 1
n = 25
Last term (l) = 25
Finding total marbles (S_25):
S_n = (n / 2) × [a + l]
S_25 = (25 / 2) × [1 + 25]
S_25 = (25 / 2) × 26
S_25 = 25 × 13
S_25 = 325
Answer: Total marbles = 325
Exercise Set 8.3 – Solutions
Class 9 Mathematics | Sequences and Progressions
Question 1
Find the \(12^{th}\) term of a GP with common ratio \(2\),
whose \(8^{th}\) term is \(192\).
Solution
We know that:
\[
t_n = ar^{\,n-1}
\]
Given:
\[
r=2
\]
\[
t_8=192
\]
For the \(8^{th}\) term:
\[
t_8=a(2)^{8-1}
\]
Therefore:
\[
192=a(2)^7
\]
Since:
\[
2^7=128
\]
So:
\[
192=128a
\]
Therefore:
\[
a=\frac{192}{128}=\frac{3}{2}
\]
Now find the \(12^{th}\) term:
\[
t_{12}=ar^{12-1}
\]
\[
t_{12}=\frac{3}{2}(2)^{11}
\]
\[
=\frac{3}{2}\times2048
\]
\[
=3072
\]
✅ Answer: The \(12^{th}\) term is
\(\boxed{3072}\).
Question 2
Find the \(10^{th}\) and \(n^{th}\) terms of the GP:
\(5,25,125,\ldots\)
Solution
Given GP:
\[
5,\;25,\;125,\ldots
\]
First term:
\[
a=5
\]
Common ratio:
\[
r=\frac{25}{5}=5
\]
Using the GP formula:
\[
t_n=ar^{\,n-1}
\]
Therefore:
\[
t_n=5(5)^{n-1}
\]
\[
t_n=5^n
\]
Now find the \(10^{th}\) term:
\[
t_{10}=5^{10}
\]
\[
t_{10}=9,765,625
\]
✅ Answer:
\(10^{th}\) term =
\(\boxed{9,765,625}\)
\(n^{th}\) term =
\(\boxed{5^n}\)
Question 3
A sequence is given by the recursive rule
\(t_1=2,\;t_{n+1}=3t_n-2\) for \(n\geq1\).
Which term of the sequence is \(730\)?
Solution
Given:
\[
t_1=2
\]
\[
t_{n+1}=3t_n-2
\]
Now calculate the terms one by one.
First term:
\[
t_1=2
\]
Second term:
\[
t_2=3t_1-2
\]
\[
=3(2)-2=4
\]
Third term:
\[
t_3=3(4)-2=10
\]
Fourth term:
\[
t_4=3(10)-2=28
\]
Fifth term:
\[
t_5=3(28)-2=82
\]
Sixth term:
\[
t_6=3(82)-2=244
\]
Seventh term:
\[
t_7=3(244)-2=730
\]
✅ Answer: \(730\) is the
\(\boxed{7^{th}}\) term.
Question 4
Which term of the GP \(2,6,18,\ldots\) is \(4374\)?
Write the explicit formula as well as the recursive formula
for the \(n^{th}\) term.
Solution
Given GP:
\[
2,\;6,\;18,\ldots
\]
First term:
\[
a=2
\]
Common ratio:
\[
r=\frac{6}{2}=3
\]
Using:
\[
t_n=ar^{\,n-1}
\]
We know that the required term is \(4374\).
\[
2(3)^{n-1}=4374
\]
Divide both sides by \(2\):
\[
3^{n-1}=2187
\]
Now:
\[
2187=3^7
\]
Therefore:
\[
3^{n-1}=3^7
\]
\[
n-1=7
\]
\[
n=8
\]
Explicit Formula
Recursive Formula
✅ Answer: \(4374\) is the
\(\boxed{8^{th}}\) term.
Question 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?
Solution
Initial height:
\[
80\text{ m}
\]
Each bounce reaches \(60\%\) of the previous height.
\[
60\%=\frac{60}{100}=0.6
\]
Therefore, the heights form a GP:
\[
80,\;80(0.6),\;80(0.6)^2,\ldots
\]
After the 5th bounce:
\[
H_5=80(0.6)^5
\]
\[
H_5=80(0.07776)
\]
\[
H_5=6.2208\text{ m}
\]
✅ Answer: After the 5th bounce, the ball reaches
\(\boxed{6.2208\text{ m}}\).
(ii) What is the total vertical distance travelled by the time
it hits the ground for the 6th time?
The first downward journey is \(80\) m.
After that, every bounce has an upward journey and a downward
journey.
The five bounce heights are:
\[
48,\;28.8,\;17.28,\;10.368,\;6.2208
\]
So the total distance is:
\[
80+2(48+28.8+17.28+10.368+6.2208)
\]
Add the bounce heights:
\[
48+28.8+17.28+10.368+6.2208
=110.6688
\]
Therefore:
\[
80+2(110.6688)
\]
\[
=80+221.3376
\]
\[
=301.3376\text{ m}
\]
✅ Answer: Total vertical distance =
\(\boxed{301.3376\text{ m}}\).
Question 6
Which term of the sequence
\(2,2\sqrt{2},4,\ldots\) is \(128\)?
Solution
Given sequence:
\[
2,\;2\sqrt{2},\;4,\ldots
\]
First term:
\[
a=2
\]
Common ratio:
\[
r=\frac{2\sqrt{2}}{2}
\]
\[
r=\sqrt{2}
\]
Using the GP formula:
\[
t_n=ar^{\,n-1}
\]
We know that \(t_n=128\).
\[
128=2(\sqrt{2})^{n-1}
\]
Divide by \(2\):
\[
64=(\sqrt{2})^{n-1}
\]
Write \(64\) as a power of \(\sqrt{2}\):
\[
64=2^6=(\sqrt{2})^{12}
\]
Therefore:
\[
(\sqrt{2})^{n-1}=(\sqrt{2})^{12}
\]
\[
n-1=12
\]
\[
n=13
\]
✅ Answer: \(128\) is the
\(\boxed{13^{th}}\) term.
Question 7
Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet.
Stage 0 is a square sheet of paper. At each stage, the square is
divided into 9 smaller squares and the centre square is removed.
The same process is repeated on the remaining 8 squares.
(i) How many red squares are there in Stages 0 to 3?
Solution
At every stage, each square produces \(8\) red squares.
| Stage |
Number of Red Squares |
| 0 |
\(1\) |
| 1 |
\(8\) |
| 2 |
\(8\times8=64\) |
| 3 |
\(64\times8=512\) |
✅ Answer: Stages 0, 1, 2 and 3 have
\(\boxed{1,\;8,\;64,\;512}\) red squares respectively.
(ii) Can you predict the number of red squares in Stages 4 and 5?
The number of squares is multiplied by \(8\) at every stage.
Stage 4:
\[
512\times8=4096
\]
Stage 5:
\[
4096\times8=32768
\]
✅ Answer:
Stage 4 = \(\boxed{4096}\) red squares
Stage 5 = \(\boxed{32768}\) red squares
(iii) Find a rule for the number of red squares at the
\(n^{th}\) stage. Write the explicit and recursive formulas.
The number of red squares is:
\[
1,\;8,\;64,\;512,\ldots
\]
This is a GP with:
\[
a=1,\qquad r=8
\]
Explicit Formula
Recursive Formula
(iv) Suppose the area of the square in Stage 0 is
1 square unit. Find the area of the red region in Stages
1, 2 and 3. Also find the areas in Stages 4 and 5.
The original square is divided into \(9\) equal squares.
One centre square is removed.
Therefore, \(8\) out of \(9\) parts remain.
\[
\text{Common ratio}=\frac{8}{9}
\]
Stage 0
\[
A_0=1
\]
Stage 1
\[
A_1=\frac{8}{9}
\]
Stage 2
\[
A_2=\left(\frac{8}{9}\right)^2
\]
\[
A_2=\frac{64}{81}
\]
Stage 3
\[
A_3=\left(\frac{8}{9}\right)^3
\]
\[
A_3=\frac{512}{729}
\]
Stage 4
\[
A_4=\left(\frac{8}{9}\right)^4
\]
\[
A_4=\frac{4096}{6561}
\]
Stage 5
\[
A_5=\left(\frac{8}{9}\right)^5
\]
\[
A_5=\frac{32768}{59049}
\]
Explicit Formula for Area
Recursive Formula for Area
What happens as \(n\) increases?
Since
\[
0<\frac{8}{9}<1,
\]
the area keeps decreasing.
As the number of stages becomes very large,
the area gets closer and closer to \(0\).
✅ Final Result:
Number of red squares:
\[
\boxed{R_n=8^n}
\]
Area of red region:
\[
\boxed{A_n=\left(\frac{8}{9}\right)^n}
\]
