Mathematics solution NCERT
Class 9 - Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions
Q1: Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Given:
T11 = 38
T16 = 73
Using Tn = a + (n − 1)d
a + 10d = 38 ..........(1)
a + 15d = 73 ..........(2)
Subtracting (1) from (2):
5d = 35
d = 7
Substituting in (1):
a + 70 = 38
a = −32
T31 = a + 30d
= −32 + 30(7)
= −32 + 210
= 178
Answer: T31 = 178
Q2: Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
Given:
3rd term = 16
7th term exceeds 5th term by 12
a + 2d = 16 ..........(1)
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
Substituting into (1):
a + 12 = 16
a = 4
AP:
4, 10, 16, 22, 28, ...
Q3: How many three-digit numbers are divisible by 7?
Smallest 3-digit multiple of 7 = 105
Largest 3-digit multiple of 7 = 994
AP: 105, 112, 119, ..., 994
a = 105, d = 7
994 = 105 + (n − 1)7
889 = 7(n − 1)
127 = n − 1
n = 128
Answer: 128 numbers
Q4: How many multiples of 4 lie between 10 and 250?
Smallest multiple = 12
Largest multiple = 248
AP: 12, 16, 20, ..., 248
248 = 12 + (n − 1)4
236 = 4(n − 1)
59 = n − 1
n = 60
Answer: 60 multiples
Q5: Find a GP for which the sum of the first two terms is – 4 and the fifth term is 4 times the third term.
First term = a
Common ratio = r
First two terms sum = −4
a + ar = −4
a(1+r) = −4 ..........(1)
5th term = 4 × 3rd term
ar4 = 4ar2
r2 = 4
r = ±2
Taking r = 2:
a(3) = −4
a = −4/3
One GP:
−4/3, −8/3, −16/3, ...
Q6: Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
100 = 18 + 19 + 20 + 21 + 22
100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16
Possible expressions:
- 18 + 19 + 20 + 21 + 22 = 100
- 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100
Q7: 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?
Initial bacteria = 30
Doubles every hour
GP: 30, 60, 120, 240, ...
a = 30
r = 2
After 2 Hours
30 × 22 = 120
After 4 Hours
30 × 24 = 480
After n Hours
T = 30 × 2n
Answers:
- 2 Hours = 120
- 4 Hours = 480
- n Hours = 30 × 2n
Q8: 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.
T4 + T8 = 24
(a+3d)+(a+7d)=24
2a+10d=24
a+5d=12 ..........(1)
T6 + T10 = 44
(a+5d)+(a+9d)=44
2a+14d=44
a+7d=22 ..........(2)
Subtracting:
2d=10
d=5
a+25=12
a=−13
First three terms:
−13, −8, −3
Answer: −13, −8, −3
Q9: Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
Sn = n(n+1)/2
n(n+1)/2 > 1000
n(n+1) > 2000
Checking values:
44 × 45 = 1980
45 × 46 = 2070
2070 > 2000
Answer: n = 45
Q10: Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.
GP: 2, 8, 32, ...
a = 2
r = 4
Tn = 2(4n−1)
131072 = 2(4n−1)
65536 = 4n−1
65536 = 216
48 = 65536
n − 1 = 8
n = 9
Answer: 131072 is the 9th term.
Explicit Formula
Tn = 2(4n−1)
Recursive Formula
T1 = 2
Tn+1 = 4Tn
Q11: 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.
Terms: a/r, a, ar
Product = −1
(a/r)(a)(ar)=a3=−1
a = −1
Sum = 13/12
−(1/r + 1 + r)=13/12
1/r + 1 + r = −13/12
12r + 12r² + 12 = −13r
12r² + 25r + 12 = 0
(3r+4)(4r+3)=0
r = −4/3 or r = −3/4
Terms:
3/4, −1, 4/3
or
4/3, −1, 3/4
Q12: If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
Let GP be:
a, ar, ar², ar³, ...
Then:
x = ar³
y = ar⁹
z = ar¹⁵
y² = (ar⁹)²
= a²r¹⁸
xz = (ar³)(ar¹⁵)
= a²r¹⁸
Therefore:
y² = xz
Hence x, y and z are in GP.
Q13: The 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.
Let terms be:
a/r, a, ar
3a = 26
a = 26/3
After solving,
r = 3/2
Terms:
52/9, 26/3, 13
Q14: Suppose 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
| n | Pn |
|---|---|
| 1 | 1 |
| 2 | 2 |
| 3 | 4 |
| 4 | 8 |
| 5 | 16 |
| 6 | 32 |
| 7 | 64 |
| 8 | 128 |
Pattern: Pn = 2n−1
Recursive: Pn = 2Pn−1
Q15:Suppose 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
| n | Wn |
|---|---|
| 1 | 1 |
| 2 | 2 |
| 3 | 5 |
| 4 | 10 |
| 5 | 20 |
| 6 | 40 |
| 7 | 80 |
| 8 | 160 |
Answer:
W1=1, W2=2, W3=5, W4=10, W5=20, W6=40, W7=80, W8=160
From W4 onwards, each term doubles.