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Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.
For the GP 2, 8, 32, ...: First term a = 2, common ratio r = 8 / 2 = 4. Explicit formula: tn = a x rⁿ⁻¹ = 2 x 4ⁿ⁻¹ Recursive formula: t1 = 2 and tn = 4 x tn-1 for n >= 2. To find which term is 131072: 2 x 4ⁿ⁻¹ = 131072 4ⁿ⁻¹ = 65536 Since 65536 = 4⁸ (as 4⁸ = 2¹⁶ = 65536): n - 1 = 8 n = 9. Hence, 1Read more
For the GP 2, 8, 32, …:
First term a = 2, common ratio r = 8 / 2 = 4.
Explicit formula:
tn = a x rⁿ⁻¹ = 2 x 4ⁿ⁻¹
Recursive formula:
t1 = 2 and tn = 4 x tn-1 for n >= 2.
To find which term is 131072:
2 x 4ⁿ⁻¹ = 131072
4ⁿ⁻¹ = 65536
Since 65536 = 4⁸ (as 4⁸ = 2¹⁶ = 65536):
n – 1 = 8
n = 9.
Hence, 131072 is the 9th term of the geometric progression.
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 smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
The sum of the first n natural numbers is given by: Sn = n(n + 1) / 2 We need the smallest n such that: n(n + 1) / 2 > 1000 n(n + 1) > 2000. Since 44 x 44 = 1936 and 45 x 45 = 2025, let us check consecutive integers near 44: For n = 44: S44 = 44 x 45 / 2 = 22 x 45 = 990 (less than 1000). For nRead more
The sum of the first n natural numbers is given by:
Sn = n(n + 1) / 2
We need the smallest n such that:
n(n + 1) / 2 > 1000
n(n + 1) > 2000.
Since 44 x 44 = 1936 and 45 x 45 = 2025,
let us check consecutive integers near 44:
For n = 44:
S44 = 44 x 45 / 2 = 22 x 45 = 990 (less than 1000).
For n = 45:
S45 = 45 x 46 / 2 = 45 x 23 = 1035 (greater than 1000).
Thus, the smallest value of n is 45.
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 lessThe 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?
The initial bacteria count is 30 and the population doubles each hour. This forms a geometric progression where the count at the end of each hour is: After 1 hour: 30 x 2 = 60 After 2 hours: 30 x 2² = 30 x 4 = 120 After 3 hours: 30 x 2³ = 240 After 4 hours: 30 x 2⁴ = 30 x 16 = 480 In general, afterRead more
The initial bacteria count is 30 and the population doubles each hour.
This forms a geometric progression where the count at the end of each hour is:
After 1 hour: 30 x 2 = 60
After 2 hours: 30 x 2² = 30 x 4 = 120
After 3 hours: 30 x 2³ = 240
After 4 hours: 30 x 2⁴ = 30 x 16 = 480
In general, after n hours:
Number of bacteria = 30 x 2ⁿ.
Hence, the counts are 120, 480 and 30 x 2ⁿ bacteria respectively.
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 lessWrite the formulae for the compounds formed by the reaction of: (i) Magnesium and nitrogen (ii) Lithium and nitrogen (iii) Sodium and sulfur (iv) Aluminium and oxygen
For magnesium and nitrogen, magnesium has a valency of 2 and nitrogen has a valency of 3, yielding magnesium nitride Mg3N2. Lithium has valency 1 and nitrogen has 3, giving lithium nitride Li3N. Sodium has valency 1 and sulfur has 2, forming sodium sulfide Na2S. Aluminium has valency 3 and oxygen haRead more
For magnesium and nitrogen, magnesium has a valency of 2 and nitrogen has a valency of 3, yielding magnesium nitride Mg3N2. Lithium has valency 1 and nitrogen has 3, giving lithium nitride Li3N. Sodium has valency 1 and sulfur has 2, forming sodium sulfide Na2S. Aluminium has valency 3 and oxygen has 2, producing aluminium oxide Al2O3. All formulae represent charge-balanced compounds.
For more NCERT Solutions of Class 9 Science Exploration Chapter 9 Atomic Foundations of Matter Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-9/
See lessComplete the Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. LiNO3 is given as an example. Cations: NH4+, Li+, Al3+, Cu2+ Anions: NO3-, SO4 2-, PO4 3-
For ammonium ion, the compounds formed are ammonium nitrate NH4NO3, ammonium sulfate (NH4)2SO4 and ammonium phosphate (NH4)3PO4. For lithium ion, the compounds are lithium nitrate LiNO3, lithium sulfate Li2SO4 and lithium phosphate Li3PO4. For aluminium ion, they are aluminium nitrate Al(NO3)3, alumRead more
For ammonium ion, the compounds formed are ammonium nitrate NH4NO3, ammonium sulfate (NH4)2SO4 and ammonium phosphate (NH4)3PO4. For lithium ion, the compounds are lithium nitrate LiNO3, lithium sulfate Li2SO4 and lithium phosphate Li3PO4. For aluminium ion, they are aluminium nitrate Al(NO3)3, aluminium sulfate Al2(SO4)3 and aluminium phosphate AlPO4. For copper(II) ion, they are copper(II) nitrate Cu(NO3)2, copper(II) sulfate CuSO4 and copper(II) phosphate Cu3(PO4)2.
For more NCERT Solutions of Class 9 Science Exploration Chapter 9 Atomic Foundations of Matter Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-9/
See less