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Sheetal Devi is shown engaging in archery using a modified technique since she lacks arms. She draws the bowstring utilizing her foot, leg and chin with sheer focus and determination. Her posture reveals intense concentration, strength and remarkable physical balance. This image clearly showcases hoRead more
Sheetal Devi is shown engaging in archery using a modified technique since she lacks arms. She draws the bowstring utilizing her foot, leg and chin with sheer focus and determination. Her posture reveals intense concentration, strength and remarkable physical balance. This image clearly showcases how human perseverance and innovative effort can overcome severe physical boundaries, transforming adversity into world-class athletic excellence on the global sporting stage.
For more NCERT Solutions of Class 9 English Kaveri Chapter 5 The World of Limitless Possibilities Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/english/kaveri-chapter-5/
See lessFind all possible ways of expressing 100 as the sum of consecutive natural numbers.
Let k consecutive natural numbers start at a, where k >= 2. Their sum is: S = (k / 2) x [2a + (k - 1)] = 100 k(2a + k - 1) = 200. Here, one factor must be odd and the other even. The odd factors of 200 are 1, 5 and 25. For k = 5: 5(2a + 4) = 200 2a + 4 = 40 2a = 36, so a = 18. For k = 25: 25(2a +Read more
Let k consecutive natural numbers start at a, where k >= 2.
Their sum is:
S = (k / 2) x [2a + (k – 1)] = 100
k(2a + k – 1) = 200.
Here, one factor must be odd and the other even. The odd factors of 200 are 1, 5 and 25.
For k = 5:
5(2a + 4) = 200
2a + 4 = 40
2a = 36, so a = 18.
For k = 25:
25(2a + 24) = 200 gives 2a + 24 = 8, so a = -8 (not natural).
Thus, the only way is 18 + 19 + 20 + 21 + 22.
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 lessDetermine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
Let the first term be a and common difference be d. Given: t7 - t5 = 12 (a + 6d) - (a + 4d) = 12 2d = 12 d = 6. Also, the third term is 16: t3 = a + 2d = 16 a + 2 x 6 = 16 a + 12 = 16 a = 4. With first term a = 4 and common difference d = 6, the terms are: First term = 4 Second term = 4 + 6 =Read more
Let the first term be a and common difference be d.
Given:
t7 – t5 = 12
(a + 6d) – (a + 4d) = 12
2d = 12
d = 6.
Also, the third term is 16:
t3 = a + 2d = 16
a + 2 x 6 = 16
a + 12 = 16
a = 4.
With first term a = 4 and common difference d = 6, the terms are:
First term = 4
Second term = 4 + 6 = 10
Third term = 10 + 6 = 16
Fourth term = 16 + 6 = 22.
Thus, the arithmetic progression is 4, 10, 16, 22, ….
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 31st term of an AP whose 11th term is 38 and 16th term is 73.
Let a be the first term and d be the common difference. Given: 11th term: t11 = a + 10d = 38 16th term: t16 = a + 15d = 73 Subtracting the first equation from the second: (a + 15d) - (a + 10d) = 73 - 38 5d = 35 d = 7. Substituting d = 7 into the first equation: a + 10 x 7 = 38 a + 70 = 38 a = 38 - 7Read more
Let a be the first term and d be the common difference.
Given:
11th term: t11 = a + 10d = 38
16th term: t16 = a + 15d = 73
Subtracting the first equation from the second:
(a + 15d) – (a + 10d) = 73 – 38
5d = 35
d = 7.
Substituting d = 7 into the first equation:
a + 10 x 7 = 38
a + 70 = 38
a = 38 – 70 = -32.
Now, finding the 31st term:
t31 = a + 30d = -32 + 30 x 7 = -32 + 210 = 178.
Hence, the 31st term is 178.
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 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 less