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Progressions

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Debojit Sarkar Tuition trainer in Jamshedpur Featured
Parsudih, Jamshedpur
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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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Enamakkal, Enamavu
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I am a school teacher from Kerala , India. I have 15 years teaching experience in CBSE school. I have a degree in statistics. I am a certified trained...

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Jyothi J. Tuition trainer in Kochi Featured
Palluruthy, Kochi
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Dwarka Sector 1, Delhi
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My name is Rajesh, and I would love to be your tutor. I have been tutoring for over 13 years. I am a former IIT student, and I received my PhD. (Computer...

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Shrimathi Aravind Tuition trainer in Bangalore Featured
Bannerghatta Main Road, Bangalore
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Venturing into the dynamic world of academia, my journey as a private tutor has been a rewarding odyssey spanning over 12 years, focusing on providing...

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Govindpuri Shiva Colony, Jaipur
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I am a teacher of mathematics. I am giving online tuitions and tuitions at my home in Jaipur(Rajasthan). I have done M.Sc. , B.Ed. in mathematics....

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Anjali A. Tuition trainer in Bangalore Featured
MSR nagar mathikere, Bangalore
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I've spent the last 30 years sharing my passion for mathematics and science with young minds in the 9th and 10th grades. Armed with a strong educational...

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Avishake Chatterjee Tuition trainer in Kolkata Featured
Behala, Kolkata
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6 yrs of Exp
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As a passionate mathematics teacher, I have always enjoyed imparting knowledge to others. I am a B.Tech graduate in Electronics and Communication....

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Ranjan kumar K Tuition trainer in Mahabubnagar Featured
Mahabubnagar, Mahabubnagar
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9 yrs of Exp
300per hour
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I am a B.Tech graduate. I completed my graduation in 2007. Because of passion for Teaching I entered in Teaching field. I have been teaching since...

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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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