The given GP has common ratio 2 and its 8th term is 192. To find the 12th term, we move four positions forward from the 8th term. Each position multiplies the previous term by 2. Therefore, t12 equals 192 multiplied by 2 raised to the fourth power. Hence, t12 equals 192 multiplied by 16, which givesRead more
The given GP has common ratio 2 and its 8th term is 192. To find the 12th term, we move four positions forward from the 8th term. Each position multiplies the previous term by 2. Therefore, t12 equals 192 multiplied by 2 raised to the fourth power. Hence, t12 equals 192 multiplied by 16, which gives 3072. Therefore, the 12th term is 3072.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
The given GP is 5, 25, 125 and so on. Here, the first term is 5 and the common ratio is 5. The nth term of a GP is tn equals a multiplied by r raised to n minus 1. Therefore, tn equals 5 multiplied by 5 raised to n minus 1, which equals 5 raised to n. Hence, the 10th term is 9765625. For moreRead more
The given GP is 5, 25, 125 and so on. Here, the first term is 5 and the common ratio is 5. The nth term of a GP is tn equals a multiplied by r raised to n minus 1. Therefore, tn equals 5 multiplied by 5 raised to n minus 1, which equals 5 raised to n. Hence, the 10th term is 9765625.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
The first term is 2. Applying the recursive rule, each new term is obtained by multiplying the previous term by 3 and subtracting 2. Thus, the sequence is 2, 4, 10, 28, 82, 244, 730. Since 730 appears at the seventh position, it is the seventh term of the sequence. Therefore, the required term is t7Read more
The first term is 2. Applying the recursive rule, each new term is obtained by multiplying the previous term by 3 and subtracting 2. Thus, the sequence is 2, 4, 10, 28, 82, 244, 730. Since 730 appears at the seventh position, it is the seventh term of the sequence. Therefore, the required term is t7, which is equal to 730.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
For the GP 2, 6, 18, the first term is 2 and the common ratio is 3. The explicit formula is tn equals 2 multiplied by 3 raised to n minus 1. Since 4374 divided by 2 equals 2187, which is 3 raised to 7, n minus 1 equals 7. Hence, n equals 8. The recursive formula is t1 equals 2 and tn equals 3tn minuRead more
For the GP 2, 6, 18, the first term is 2 and the common ratio is 3. The explicit formula is tn equals 2 multiplied by 3 raised to n minus 1. Since 4374 divided by 2 equals 2187, which is 3 raised to 7, n minus 1 equals 7. Hence, n equals 8. The recursive formula is t1 equals 2 and tn equals 3tn minus 1.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
(i) Answer: After the first bounce, the ball reaches 60% of 80 metres, which is 48 metres. The successive bounce heights form a GP with first term 48 and common ratio 0.6. Therefore, the height after the fifth bounce is 48 multiplied by 0.6 raised to 4. This equals 6.2208 metres. Hence, the ball reaRead more
(i) Answer:
After the first bounce, the ball reaches 60% of 80 metres, which is 48 metres. The successive bounce heights form a GP with first term 48 and common ratio 0.6. Therefore, the height after the fifth bounce is 48 multiplied by 0.6 raised to 4. This equals 6.2208 metres. Hence, the ball reaches a height of 6.2208 metres after the fifth bounce.
(ii) Answer:
The ball first travels 80 metres downward. Before hitting the ground for the sixth time, it rises and falls five times. The five heights are 48, 28.8, 17.28, 10.368 and 6.2208 metres. Their sum is 110.6688 metres. Therefore, total vertical distance equals 80 plus twice 110.6688, which is 301.3376 metres. Hence, the ball has travelled 301.3376 metres when it hits the ground for the sixth time.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
The given GP has common ratio 2 and its 8th term is 192. To find the 12th term, we move four positions forward from the 8th term. Each position multiplies the previous term by 2. Therefore, t12 equals 192 multiplied by 2 raised to the fourth power. Hence, t12 equals 192 multiplied by 16, which givesRead more
The given GP has common ratio 2 and its 8th term is 192. To find the 12th term, we move four positions forward from the 8th term. Each position multiplies the previous term by 2. Therefore, t12 equals 192 multiplied by 2 raised to the fourth power. Hence, t12 equals 192 multiplied by 16, which gives 3072. Therefore, the 12th term is 3072.
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 10th and nth terms of the GP: 5, 25, 125, …
The given GP is 5, 25, 125 and so on. Here, the first term is 5 and the common ratio is 5. The nth term of a GP is tn equals a multiplied by r raised to n minus 1. Therefore, tn equals 5 multiplied by 5 raised to n minus 1, which equals 5 raised to n. Hence, the 10th term is 9765625. For moreRead more
The given GP is 5, 25, 125 and so on. Here, the first term is 5 and the common ratio is 5. The nth term of a GP is tn equals a multiplied by r raised to n minus 1. Therefore, tn equals 5 multiplied by 5 raised to n minus 1, which equals 5 raised to n. Hence, the 10th term is 9765625.
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 lessA sequence is given by the recursive rule t1 = 2, tn+1 = 3tn minus 2 for n greater than or equal to 1. Which term of the sequence is 730?
The first term is 2. Applying the recursive rule, each new term is obtained by multiplying the previous term by 3 and subtracting 2. Thus, the sequence is 2, 4, 10, 28, 82, 244, 730. Since 730 appears at the seventh position, it is the seventh term of the sequence. Therefore, the required term is t7Read more
The first term is 2. Applying the recursive rule, each new term is obtained by multiplying the previous term by 3 and subtracting 2. Thus, the sequence is 2, 4, 10, 28, 82, 244, 730. Since 730 appears at the seventh position, it is the seventh term of the sequence. Therefore, the required term is t7, which is equal to 730.
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 lessWhich term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.
For the GP 2, 6, 18, the first term is 2 and the common ratio is 3. The explicit formula is tn equals 2 multiplied by 3 raised to n minus 1. Since 4374 divided by 2 equals 2187, which is 3 raised to 7, n minus 1 equals 7. Hence, n equals 8. The recursive formula is t1 equals 2 and tn equals 3tn minuRead more
For the GP 2, 6, 18, the first term is 2 and the common ratio is 3. The explicit formula is tn equals 2 multiplied by 3 raised to n minus 1. Since 4374 divided by 2 equals 2187, which is 3 raised to 7, n minus 1 equals 7. Hence, n equals 8. The recursive formula is t1 equals 2 and tn equals 3tn minus 1.
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 lessA 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?
(i) Answer: After the first bounce, the ball reaches 60% of 80 metres, which is 48 metres. The successive bounce heights form a GP with first term 48 and common ratio 0.6. Therefore, the height after the fifth bounce is 48 multiplied by 0.6 raised to 4. This equals 6.2208 metres. Hence, the ball reaRead more
(i) Answer:
After the first bounce, the ball reaches 60% of 80 metres, which is 48 metres. The successive bounce heights form a GP with first term 48 and common ratio 0.6. Therefore, the height after the fifth bounce is 48 multiplied by 0.6 raised to 4. This equals 6.2208 metres. Hence, the ball reaches a height of 6.2208 metres after the fifth bounce.
(ii) Answer:
The ball first travels 80 metres downward. Before hitting the ground for the sixth time, it rises and falls five times. The five heights are 48, 28.8, 17.28, 10.368 and 6.2208 metres. Their sum is 110.6688 metres. Therefore, total vertical distance equals 80 plus twice 110.6688, which is 301.3376 metres. Hence, the ball has travelled 301.3376 metres when it hits the ground for the sixth time.
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