Question 1

Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

Accepted Answer

Given: Common ratio r = 2, 8th term t_8 = 192.

Formula for nth term of GP: t_n = a * r^(n-1)

So, t_8 = a * 2^(8-1) = a * 2^7 = a * 128 = 192
=> a = 192 / 128 = 1.5

Now, find 12th term:
t_12 = a * 2^(12-1) = 1.5 * 2^11 = 1.5 * 2048 = 3072

Answer: The 12th term is 3072. — Step 1: Use the formula t_n = a * r^(n-1).
Step 2: Substitute t_8 = 192 and r = 2 to find a.
Step 3: Calculate a = 192 / 128 = 1.5.
Step 4: Use a to find t_12 = 1.5 * 2^11 = 3072.

Question 2

Find the 10th and nth terms of the GP: 5, 25, 125, ...

Accepted Answer

Given GP: 5, 25, 125, ...
First term a = 5
Common ratio r = 25 / 5 = 5

10th term:
t_10 = a * r^(10-1) = 5 * 5^9 = 5 * 1953125 = 9765625

nth term:
t_n = a * r^{n-1} = 5 * 5^{n-1} = 5^n

Answer:
10th term = 9765625
nth term = 5^n — Step 1: Identify a = 5 and r = 5.
Step 2: Use formula t_n = a * r^{n-1}.
Step 3: Calculate t_10 = 5 * 5^9 = 9765625.
Step 4: Express nth term as t_n = 5^n.

Question 3

A sequence is given by the recursive rule t_1 = 2, t_{n+1} = 3t_n - 2 for n ≥ 1. Which term of the sequence is 730?

Accepted Answer

Given: t_1 = 2, t_{n+1} = 3t_n - 2

We want to find n such that t_n = 730.

First, solve the recurrence:
Assume t_n = A * 3^{n-1} + B (particular solution)

Substitute into recurrence:
t_{n+1} = 3 t_n - 2
=> A * 3^n + B = 3 (A * 3^{n-1} + B) - 2
=> A * 3^n + B = A * 3^n + 3B - 2

Equate constants:
B = 3B - 2 => 2B = 2 => B = 1

Use initial condition t_1 = 2:
2 = A * 3^{0} + 1 => A + 1 = 2 => A = 1

So, t_n = 3^{n-1} + 1

Set t_n = 730:
3^{n-1} + 1 = 730 => 3^{n-1} = 729

Since 729 = 3^6,

n - 1 = 6 => n = 7

Answer: The 7th term is 730. — Step 1: Assume solution t_n = A * 3^{n-1} + B.
Step 2: Substitute into recurrence to find B.
Step 3: Use initial condition to find A.
Step 4: Write explicit formula t_n = 3^{n-1} + 1.
Step 5: Solve 3^{n-1} + 1 = 730 to find n = 7.

Question 4

Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the nth term.

Accepted Answer

Given GP: 2, 6, 18, ...
First term a = 2
Common ratio r = 6 / 2 = 3

Find n such that t_n = 4374

Formula: t_n = a * r^{n-1} = 2 * 3^{n-1}

Set equal:
2 * 3^{n-1} = 4374
=> 3^{n-1} = 4374 / 2 = 2187

Note 2187 = 3^7

So, n - 1 = 7 => n = 8

Explicit formula: t_n = 2 * 3^{n-1}
Recursive formula: t_1 = 2, t_{n} = 3 * t_{n-1}

Answer:
The 8th term is 4374.
Explicit formula: t_n = 2 * 3^{n-1}
Recursive formula: t_1 = 2, t_n = 3 t_{n-1} — Step 1: Identify a = 2, r = 3.
Step 2: Use formula t_n = a r^{n-1}.
Step 3: Solve 2 * 3^{n-1} = 4374.
Step 4: Find n = 8.
Step 5: Write explicit and recursive formulas.

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?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?

Accepted Answer

(i) Height after 5th bounce:

Initial height h_0 = 80 m
Bounce heights form a GP with first term h_1 = 0.6 * 80 = 48 m
Common ratio r = 0.6

Height after 5th bounce:
h_5 = h_1 * r^{5-1} = 48 * 0.6^4
Calculate 0.6^4 = 0.1296
h_5 = 48 * 0.1296 = 6.2208 m

Answer: The ball reaches approximately 6.22 metres after the 5th bounce.

(ii) Total vertical distance travelled by the time it hits the ground for the 6th time:

The ball travels down 80 m initially.
Then it bounces up 48 m, down 48 m, up 28.8 m, down 28.8 m, and so on.

Total distance = initial drop + 2 * sum of first 5 bounce heights
Sum of first 5 bounce heights = h_1 + h_2 + h_3 + h_4 + h_5
= 48 + 48*0.6 + 48*0.6^2 + 48*0.6^3 + 48*0.6^4
= 48 * (1 + 0.6 + 0.36 + 0.216 + 0.1296)
= 48 * 2.3056 = 110.6688 m

Total distance = 80 + 2 * 110.6688 = 80 + 221.3376 = 301.3376 m

Answer: The total vertical distance travelled by the time it hits the ground for the 6th time is approximately 301.34 metres. — Part (i):
- The bounce heights form a GP with first term 48 m and ratio 0.6.
- Use formula for nth term: h_n = h_1 * r^{n-1}.
- Calculate h_5 = 48 * 0.6^4 = 6.2208 m.

Part (ii):
- Total distance = initial drop + twice the sum of first 5 bounce heights.
- Sum of GP terms: S_5 = h_1 * (1 - r^5) / (1 - r) = 48 * (1 - 0.6^5) / 0.4.
- Calculate sum and total distance accordingly.

Question 6

Which term of the sequence 2, 2√2, 4, ... is 128?

Accepted Answer

Given sequence: 2, 2√2, 4, ...

Check if this is a GP:
Ratio r = (2√2) / 2 = √2

First term a = 2

General term:
t_n = a * r^{n-1} = 2 * (\sqrt{2})^{n-1}

We want t_n = 128

So,
2 * (\sqrt{2})^{n-1} = 128
=> (\sqrt{2})^{n-1} = 64

Note that \sqrt{2} = 2^{1/2}
So,
(2^{1/2})^{n-1} = 2^{(n-1)/2} = 64 = 2^6

Equate exponents:
(n-1)/2 = 6 => n - 1 = 12 => n = 13

Answer: The 13th term is 128. — Step 1: Identify a = 2 and r = √2.
Step 2: Use formula t_n = a r^{n-1}.
Step 3: Set t_n = 128 and solve for n.
Step 4: Convert √2 to 2^{1/2} and equate powers.
Step 5: Find n = 13.

Question 7

Fig. 8.12 shows Stages 0 to 3 of the Sierpinski 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.

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?

Accepted Answer

(i) Number of red squares:
Stage 0: 1 (the whole square)
Stage 1: 8 (since center square removed, 9 - 1 = 8)
Stage 2: 8 * 8 = 8^2 = 64
Stage 3: 8^3 = 512

(ii) Number of red squares at Stage 4 = 8^4 = 4096
Number of red squares at Stage 5 = 8^5 = 32768

(iii) Rule:
Number of red squares at nth stage = 8^n
Explicit formula: N_n = 8^n
Recursive formula: N_0 = 1, N_n = 8 * N_{n-1}

(iv) Area of red region:
Area at Stage 0 = 1 sq unit
At each stage, each red square is divided into 9 smaller squares and the center one is removed, so area reduces by factor 8/9.

Area at Stage 1 = (8/9) * 1 = 8/9
Area at Stage 2 = (8/9)^2
Area at Stage 3 = (8/9)^3

Area at Stage n = (8/9)^n

Explicit formula: A_n = (8/9)^n
Recursive formula: A_0 = 1, A_n = (8/9) * A_{n-1}

As n increases, (8/9)^n approaches 0, so the red area tends to 0.

Answer: The number of red squares grows exponentially as 8^n, and the red area shrinks exponentially approaching zero as n increases. — Step 1: Count red squares at each stage using the fractal construction.
Step 2: Identify the pattern as powers of 8.
Step 3: Write explicit and recursive formulas for number of red squares.
Step 4: Calculate area reduction factor 8/9 at each stage.
Step 5: Write explicit and recursive formulas for area.
Step 6: Conclude area tends to zero as n increases.

Question 8

Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Accepted Answer

Let the first term be a and common difference be d.

Given: t_11 = a + 10d = 38 ...(1)
       t_16 = a + 15d = 73 ...(2)

Subtract (1) from (2):
(a + 15d) - (a + 10d) = 73 - 38
5d = 35
=> d = 7

Substitute d = 7 in (1):
a + 10*7 = 38
=> a + 70 = 38
=> a = 38 - 70 = -32

Now, find t_31:
t_31 = a + 30d = -32 + 30*7 = -32 + 210 = 178

Therefore, the 31st term is 178. — We use the formula for the nth term of an AP: t_n = a + (n-1)d.
Given two terms, we form two equations and solve for a and d.
Then substitute values to find the required term.

Question 9

Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

Accepted Answer

Let the first term be a and common difference be d.

Given:
t_3 = a + 2d = 16 ...(1)

Also, t_7 - t_5 = 12
=> (a + 6d) - (a + 4d) = 12
=> 2d = 12
=> d = 6

Substitute d = 6 in (1):
a + 2*6 = 16
=> a + 12 = 16
=> a = 4

Therefore, the AP is: 4, 10, 16, 22, 28, 34, 40, ... — Using the nth term formula t_n = a + (n-1)d, we set up equations from given terms and solve for a and d.
Then write the AP using these values.

Question 10

Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Accepted Answer

We want to express 100 as sum of k consecutive natural numbers starting from m:

Sum = m + (m+1) + (m+2) + ... + (m + k - 1) = 100

Sum of k terms starting from m:
S = k/2 * [2m + (k - 1)] = 100

Rearranged:
2m + k - 1 = 200/k

Since m and k are natural numbers, 200/k - (k - 1) must be even and positive.

Try values of k such that 200/k is integer:
k = 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200

Check each k:

k=5:
2m + 4 = 200/5 = 40
2m = 36
m=18 (valid)
Sum: 18+19+20+21+22=100

k=8:
2m + 7 = 200/8=25
2m=18
m=9 (valid)
Sum: 9+10+11+12+13+14+15+16=100

k=1:
2m + 0 = 200/1=200
2m=200
m=100 (valid)
Sum: 100

k=2:
2m + 1=100
2m=99
m=49.5 (not natural)

k=4:
2m + 3=50
2m=47
m=23.5 (no)

k=10:
2m + 9=20
2m=11
m=5.5 (no)

k=20:
2m + 19=10
2m=-9 (no)

k=25:
2m + 24=8
2m=-16 (no)

k=40:
2m + 39=5
2m=-34 (no)

k=50:
2m + 49=4
2m=-45 (no)

k=100:
2m + 99=2
2m=-97 (no)

k=200:
2m + 199=1
2m=-198 (no)

Therefore, the possible sequences are:
- 100 alone (k=1, m=100)
- 18 to 22 (k=5, m=18)
- 9 to 16 (k=8, m=9)

Hence, 100 can be expressed as sum of consecutive natural numbers in these three ways. — We use the formula for sum of k consecutive natural numbers starting from m.
By trying divisors of 200 and checking for natural number solutions, we find valid sequences.

Question 11

The 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?

Accepted Answer

Given initial bacteria count a = 30.
The bacteria double every hour, so common ratio r = 2.

Number of bacteria after n hours is given by the nth term of a GP:

T_n = a * r^{n-1}

At the end of 2nd hour:
T_2 = 30 * 2^{2-1} = 30 * 2 = 60

At the end of 4th hour:
T_4 = 30 * 2^{4-1} = 30 * 8 = 240

At the end of nth hour:
T_n = 30 * 2^{n-1}

Therefore, bacteria counts are 60 at 2nd hour, 240 at 4th hour, and 30*2^{n-1} at nth hour. — Since the bacteria count doubles every hour, the sequence forms a geometric progression with first term 30 and common ratio 2.
We use the formula for nth term of GP to find counts at specified hours.

Question 12

The 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.

Accepted Answer

Let the first term be a and common difference be d.

Given:
t_4 + t_8 = 24
=> (a + 3d) + (a + 7d) = 24
=> 2a + 10d = 24 ...(1)

Also,
t_6 + t_10 = 44
=> (a + 5d) + (a + 9d) = 44
=> 2a + 14d = 44 ...(2)

Subtract (1) from (2):
(2a + 14d) - (2a + 10d) = 44 - 24
4d = 20
=> d = 5

Substitute d=5 in (1):
2a + 10*5 = 24
2a + 50 = 24
2a = 24 - 50 = -26
=> a = -13

First three terms:
t_1 = a = -13
 t_2 = a + d = -13 + 5 = -8
 t_3 = a + 2d = -13 + 10 = -3

Therefore, the first three terms are -13, -8, -3. — Using the nth term formula t_n = a + (n-1)d, we form two equations from given sums and solve for a and d.
Then calculate the first three terms.

Question 13

Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Accepted Answer

Sum of first n natural numbers is given by:
S_n = n(n+1)/2

We want S_n > 1000
=> n(n+1)/2 > 1000
=> n(n+1) > 2000

Try to find smallest integer n satisfying this:

Check n=44:
44*45 = 1980 < 2000 (No)

Check n=45:
45*46 = 2070 > 2000 (Yes)

Therefore, smallest n is 45. — We use the formula for sum of first n natural numbers and solve inequality to find smallest n.
Trial with integers near sqrt(2000) helps find exact n.

Question 14

Which term of the GP: 2, 8, 32, ... is 131072? Write the explicit formula as well as the recursive formula for the nth term.

Accepted Answer

Given GP: 2, 8, 32, ...
First term a = 2
Common ratio r = 8/2 = 4

General term:
T_n = a * r^{n-1} = 2 * 4^{n-1}

Find n such that T_n = 131072

2 * 4^{n-1} = 131072
=> 4^{n-1} = 131072 / 2 = 65536

Note: 4 = 2^2
So, 4^{n-1} = (2^2)^{n-1} = 2^{2(n-1)}

65536 = 2^{16}

Therefore,
2^{2(n-1)} = 2^{16}
=> 2(n-1) = 16
=> n - 1 = 8
=> n = 9

Hence, 131072 is the 9th term.

Explicit formula: T_n = 2 * 4^{n-1}
Recursive formula: T_1 = 2; T_n = 4 * T_{n-1} for n > 1. — We use the nth term formula for GP and solve for n by equating the term to 131072.
Express powers of 4 in terms of powers of 2 to solve the exponent equation.
Write explicit and recursive formulas accordingly.

Question 15

The sum of the first three terms of a GP is 13/12 and their product is −1. Find the common ratio and the terms.

Accepted Answer

Let the first term be a and common ratio be r.

Given:
S = a + ar + ar^2 = 13/12 ...(1)
Product = a * ar * ar^2 = a^3 r^3 = -1 ...(2)

From (2):
(a r)^3 = -1
=> a r = -1

From (1):
a + ar + ar^2 = a (1 + r + r^2) = 13/12

Substitute a = -1/r from a r = -1:
(-1/r)(1 + r + r^2) = 13/12
=> -(1 + r + r^2)/r = 13/12
=> -(1 + r + r^2) = (13/12) r
=> Multiply both sides by 12:
-12 - 12r - 12r^2 = 13 r
=> -12 - 12r - 12r^2 - 13r = 0
=> -12 - 25r - 12r^2 = 0
Multiply both sides by -1:
12 + 25r + 12r^2 = 0

Rewrite:
12r^2 + 25r + 12 = 0

Solve quadratic:
Discriminant D = 25^2 - 4*12*12 = 625 - 576 = 49

r = [-25 ± 7]/(2*12) = [-25 ± 7]/24

Two roots:
r_1 = (-25 + 7)/24 = -18/24 = -3/4
r_2 = (-25 - 7)/24 = -32/24 = -4/3

Find a for each r:
For r = -3/4:
a r = -1 => a = -1 / r = -1 / (-3/4) = 4/3
Terms: a = 4/3, ar = 4/3 * (-3/4) = -1, ar^2 = 4/3 * (9/16) = 3/4

For r = -4/3:
a = -1 / (-4/3) = 3/4
Terms: a = 3/4, ar = 3/4 * (-4/3) = -1, ar^2 = 3/4 * (16/9) = 4/3

Therefore, the two possible sets of terms are:
(4/3, -1, 3/4) and (3/4, -1, 4/3). — Using sum and product formulas for GP terms, we form equations and solve for r using quadratic formula.
Then find a and write terms accordingly.

Predicting What Comes Next: Exploring Sequences and

Predicting What Comes Next: Exploring Sequences and — Study Notes

Introduction

Arithmetic Progression

Sum of First n Terms of an Arithmetic Progression

Practice Questions — Predicting What Comes Next: Exploring Sequences and

All 8 Chapters in Ganita Manjari (English)