For this AP, the first term is a = 21 and the common difference is d = 18 - 21 = -3. The nth term formula is tn = a + (n - 1)d. For tn = -81: 21 + (n - 1) x (-3) = -81 (n - 1) x (-3) = -102 n - 1 = 34, so n = 35. Thus, -81 is the 35th term. For tn = 0: 21 + (n - 1) x (-3) = 0 (n - 1) x (-3) = -21 nRead more
For this AP, the first term is a = 21 and the common difference is d = 18 – 21 = -3.
The nth term formula is tn = a + (n – 1)d.
For tn = -81:
21 + (n – 1) x (-3) = -81
(n – 1) x (-3) = -102
n – 1 = 34, so n = 35. Thus, -81 is the 35th term.
For tn = 0:
21 + (n – 1) x (-3) = 0
(n – 1) x (-3) = -21
n – 1 = 7, so n = 8.
Since n is a natural number, 0 is the 8th term.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
In the given sequence, the first term is a = 11 and the common difference is d = 8 - 11 = -3. The explicit rule for the nth term is: tn = a + (n - 1)d tn = 11 + (n - 1) x (-3) tn = 11 - 3n + 3 tn = 14 - 3n. To write the recursive rule, each term after the first is obtained by adding the common diffeRead more
In the given sequence, the first term is a = 11 and the common difference is d = 8 – 11 = -3.
The explicit rule for the nth term is:
tn = a + (n – 1)d
tn = 11 + (n – 1) x (-3)
tn = 11 – 3n + 3
tn = 14 – 3n.
To write the recursive rule, each term after the first is obtained by adding the common difference:
t1 = 11 and tn = tn-1 – 3 for n >= 2.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
Given an AP of 50 terms: 3rd term: a + 2d = 12 50th term: a + 49d = 106 Subtracting the first equation from the second: (a + 49d) - (a + 2d) = 106 - 12 47d = 94, which gives d = 2. Substitute d = 2 into the first equation: a + 2 x 2 = 12 a = 12 - 4 = 8. Now, finding the 29th term: t29 = a + 28d = 8Read more
Given an AP of 50 terms:
3rd term: a + 2d = 12
50th term: a + 49d = 106
Subtracting the first equation from the second:
(a + 49d) – (a + 2d) = 106 – 12
47d = 94, which gives d = 2.
Substitute d = 2 into the first equation:
a + 2 x 2 = 12
a = 12 – 4 = 8.
Now, finding the 29th term:
t29 = a + 28d = 8 + 28 x 2 = 8 + 56 = 64.
Hence, the 29th term is 64.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
The two-digit numbers divisible by 3 form the sequence: 12, 15, 18, ..., 99. Here, a = 12, d = 3 and last term tn = 99. Using tn = a + (n - 1)d: 99 = 12 + (n - 1) x 3 87 = (n - 1) x 3 n - 1 = 29, so n = 30. There are 30 two-digit numbers divisible by 3. Their sum is given by Sn = (n / 2) x (a + tn):Read more
The two-digit numbers divisible by 3 form the sequence: 12, 15, 18, …, 99.
Here, a = 12, d = 3 and last term tn = 99.
Using tn = a + (n – 1)d:
99 = 12 + (n – 1) x 3
87 = (n – 1) x 3
n – 1 = 29, so n = 30.
There are 30 two-digit numbers divisible by 3.
Their sum is given by Sn = (n / 2) x (a + tn):
S30 = (30 / 2) x (12 + 99) = 15 x 111 = 1665.
Thus, the sum is 1665.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
Harish's yearly income forms an arithmetic progression where: First term a = 500000 Common difference d = 20000 Target term tn = 700000. Using the nth term formula tn = a + (n - 1)d: 700000 = 500000 + (n - 1) x 20000 200000 = (n - 1) x 20000 n - 1 = 10 n = 11. Since n = 11 represents the 11th year oRead more
Harish’s yearly income forms an arithmetic progression where:
First term a = 500000
Common difference d = 20000
Target term tn = 700000.
Using the nth term formula tn = a + (n – 1)d:
700000 = 500000 + (n – 1) x 20000
200000 = (n – 1) x 20000
n – 1 = 10
n = 11.
Since n = 11 represents the 11th year of his career, it required 10 annual increments.
Therefore, his income reached 7,00,000 after 10 years.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer.
For this AP, the first term is a = 21 and the common difference is d = 18 - 21 = -3. The nth term formula is tn = a + (n - 1)d. For tn = -81: 21 + (n - 1) x (-3) = -81 (n - 1) x (-3) = -102 n - 1 = 34, so n = 35. Thus, -81 is the 35th term. For tn = 0: 21 + (n - 1) x (-3) = 0 (n - 1) x (-3) = -21 nRead more
For this AP, the first term is a = 21 and the common difference is d = 18 – 21 = -3.
The nth term formula is tn = a + (n – 1)d.
For tn = -81:
21 + (n – 1) x (-3) = -81
(n – 1) x (-3) = -102
n – 1 = 34, so n = 35. Thus, -81 is the 35th term.
For tn = 0:
21 + (n – 1) x (-3) = 0
(n – 1) x (-3) = -21
n – 1 = 7, so n = 8.
Since n is a natural number, 0 is the 8th term.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See lessFind the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
In the given sequence, the first term is a = 11 and the common difference is d = 8 - 11 = -3. The explicit rule for the nth term is: tn = a + (n - 1)d tn = 11 + (n - 1) x (-3) tn = 11 - 3n + 3 tn = 14 - 3n. To write the recursive rule, each term after the first is obtained by adding the common diffeRead more
In the given sequence, the first term is a = 11 and the common difference is d = 8 – 11 = -3.
The explicit rule for the nth term is:
tn = a + (n – 1)d
tn = 11 + (n – 1) x (-3)
tn = 11 – 3n + 3
tn = 14 – 3n.
To write the recursive rule, each term after the first is obtained by adding the common difference:
t1 = 11 and tn = tn-1 – 3 for n >= 2.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See lessAn AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Given an AP of 50 terms: 3rd term: a + 2d = 12 50th term: a + 49d = 106 Subtracting the first equation from the second: (a + 49d) - (a + 2d) = 106 - 12 47d = 94, which gives d = 2. Substitute d = 2 into the first equation: a + 2 x 2 = 12 a = 12 - 4 = 8. Now, finding the 29th term: t29 = a + 28d = 8Read more
Given an AP of 50 terms:
3rd term: a + 2d = 12
50th term: a + 49d = 106
Subtracting the first equation from the second:
(a + 49d) – (a + 2d) = 106 – 12
47d = 94, which gives d = 2.
Substitute d = 2 into the first equation:
a + 2 x 2 = 12
a = 12 – 4 = 8.
Now, finding the 29th term:
t29 = a + 28d = 8 + 28 x 2 = 8 + 56 = 64.
Hence, the 29th term is 64.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See lessHow many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
The two-digit numbers divisible by 3 form the sequence: 12, 15, 18, ..., 99. Here, a = 12, d = 3 and last term tn = 99. Using tn = a + (n - 1)d: 99 = 12 + (n - 1) x 3 87 = (n - 1) x 3 n - 1 = 29, so n = 30. There are 30 two-digit numbers divisible by 3. Their sum is given by Sn = (n / 2) x (a + tn):Read more
The two-digit numbers divisible by 3 form the sequence: 12, 15, 18, …, 99.
Here, a = 12, d = 3 and last term tn = 99.
Using tn = a + (n – 1)d:
99 = 12 + (n – 1) x 3
87 = (n – 1) x 3
n – 1 = 29, so n = 30.
There are 30 two-digit numbers divisible by 3.
Their sum is given by Sn = (n / 2) x (a + tn):
S30 = (30 / 2) x (12 + 99) = 15 x 111 = 1665.
Thus, the sum is 1665.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See lessHarish 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?
Harish's yearly income forms an arithmetic progression where: First term a = 500000 Common difference d = 20000 Target term tn = 700000. Using the nth term formula tn = a + (n - 1)d: 700000 = 500000 + (n - 1) x 20000 200000 = (n - 1) x 20000 n - 1 = 10 n = 11. Since n = 11 represents the 11th year oRead more
Harish’s yearly income forms an arithmetic progression where:
First term a = 500000
Common difference d = 20000
Target term tn = 700000.
Using the nth term formula tn = a + (n – 1)d:
700000 = 500000 + (n – 1) x 20000
200000 = (n – 1) x 20000
n – 1 = 10
n = 11.
Since n = 11 represents the 11th year of his career, it required 10 annual increments.
Therefore, his income reached 7,00,000 after 10 years.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/
See less