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Progressions

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1
Debojit Sarkar Tuition trainer in Jamshedpur Featured
Parsudih, Jamshedpur
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3 yrs of Exp
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I am a B.E(Civil) from Jadavpur University. I am teaching since 2021. I am very proficient in Maths and Science. My main goal is to make students...

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Prem Nagar, Hisar
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I recently retired from my job after serving in a very reputed school. I have good experience in teaching accountancy and business studies. I know...

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Thane West, Thane
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With over 15 years of dedicated experience in teaching mathematics, science, and computer science, I have cultivated a deep understanding and love...

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Enamakkal, Enamavu
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Rajeev Kumar Giri Tuition trainer in Varanasi Featured
Pandeypur, Varanasi
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10 yrs of Exp
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I am a CBSE School teacher since 2011 teaching IX to XII. I am teaching Mathematics. Some of my students are teaching studying in different IITs....

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Girija Srinarayan Tuition trainer in Bangalore Featured
HMT Estate 2nd Stage, Bangalore
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I am a teacher. Teaching chemistry for the past 20+ years. I also have some experience in research in applied chemistry. I have a masters degree in...

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Ashok Nagar D' Souza Layout, Bangalore
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With over 10 years of teaching experience across various educational systems, including IGCSE, ICSE, and CBSE, I have built a strong foundation in...

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Seema Nagwan Tuition trainer in Delhi Featured
Siraspur, Delhi
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Indrajeet K. Tuition trainer in Noida Featured
Sector 1 Greater Noida, Noida
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I completed my B.Tech in computer technology and MBA in marketing. I have six years of experience. I began giving home tuition in 2016 in Mumbai and...

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Pushpendra Sharma Tuition trainer in Delhi Featured
Mahavir Nagar, Delhi
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12 yrs of Exp
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Progressions Lessons

Geometric Progression
If a is the first term and r is the common ratio of the geometric progression, then its nth term is given by an = arn-1 The sum Sn of the first n terms of the G.P. is given by Sn = a (rn – 1)/...

Notes on Harmonic progression
Let a, b and c form an H.P. Then 1/a, 1/b and 1/c form an A.P. If a, b and c are in H.P. then 2/b = 1/a + 1/c, which can be simplified as b = 2ac/(a+c) If a and b are two non-zero numbers...

Geometric Progression : Basics
Geometric Mean (GM) The geometric mean is similar to the arithmetic mean, but it uses the property of multiplication, to find the mean between any two numbers. GM of two numbers a and b may be defined...

Progressions Questions

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Answered on 30/11/2016 Learn Progressions +2 Tuition/Class IX-X Tuition

Tapas Bhattacharya

Tutor

The first term, t1=a = 7. The common difference, b=13 - 7=6. Last term or the nth term, tn=a + (n-1)b = 205. So, we have, 7 + (n - 1)6 = 205 --> (n - 1) = (205 - 7)/6 = 33 --> n = 34. Answer: Number of terms is 34.
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Answered on 01/12/2016 Learn Progressions +2 Tuition/Class IX-X Tuition

Tapas Bhattacharya

Tutor

Let the first term be a and the common difference be b. So, the 10th term: t10 = a + 9b = 5. and, the 18th term: t18 = a + 17b = 77. So, from t17 - t10 we get: 8b = 77 - 5 = 72. This gives: b = 9. From t10 we get: a = 5 - 9b = 5 - 9x9 = -76. So, a = -76, b = 9 and the A.P is:- -76, -67, -58, -49,... read more
Let the first term be a and the common difference be b. So, the 10th term: t10 = a + 9b = 5. and, the 18th term: t18 = a + 17b = 77. So, from t17 - t10 we get: 8b = 77 - 5 = 72. This gives: b = 9. From t10 we get: a = 5 - 9b = 5 - 9x9 = -76. So, a = -76, b = 9 and the A.P is:- -76, -67, -58, -49, ..... (Answer). read less
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Answered on 01/12/2016 Learn Progressions +2 Tuition/Class IX-X Tuition

Tapas Bhattacharya

Tutor

Let the first term be (a - 5b) and the common difference be 2b. So the six terms are: (a - 5b), (a - 3b), (a - b), (a + b), (a + 3b), (a + 5b). By the given conditions:- Sum of six terms = (a - 5b) + (a - 3b) + (a - b) + (a + b) + (a + 3b) + (a + 5b) = 6a = 0. So, a = 0. The fourth term = a + b = 2... read more
Let the first term be (a - 5b) and the common difference be 2b. So the six terms are: (a - 5b), (a - 3b), (a - b), (a + b), (a + 3b), (a + 5b). By the given conditions:- Sum of six terms=(a - 5b) + (a - 3b) + (a - b) + (a + b) + (a + 3b) + (a + 5b) = 6a=0. So, a=0. The fourth term=a + b=2 --> 0 + b=2 --> b = 2. So, for the A.P. the first term x = (a - 5b) = -10. The common difference, y = 2b = 4. Sum of 1st 30 terms, S30 = (n/2)[2x + (n-1)y] = 15 x [ -20 + 29 x 4] = 1440. (Answer) read less
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Answered on 01/12/2016 Learn Progressions +2 Tuition/Class IX-X Tuition

Tapas Bhattacharya

Tutor

Let the first term be a and the common difference be b. So, t1 = a = 7 and t60 = a + 59b = 125. From above, we get:- t60 - t1 = 59b = 125 - 7 = 118. So, b = 118/59 = 2. Hence, the 32nd term, t32 = a + 31b = =7 + 31x2 = 69. (Answer)
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Answered on 01/12/2016 Learn Progressions +2 Tuition/Class IX-X Tuition

Tapas Bhattacharya

Tutor

The first four terms are 10, 10 + 10, 10 + 2x10, and 10 + 3x10. So, the terms are: 10, 20, 30, 40.
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