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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 CBSE - Class 10/Mathematics

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 CBSE - Class 10/Mathematics

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 CBSE - Class 10/Mathematics

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 CBSE - Class 10/Mathematics

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 CBSE - Class 10/Mathematics

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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