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Progressions

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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)/...

Raj Kumar

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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...

Raj Kumar

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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...

Hemant Pandey

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Progressions Questions

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Answered on 30/11/2016 Learn Progressions +2 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Find the number of terms of the A.P: 7, 13, 19, ...., 205?

Tapas Bhattacharya

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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 Tuition/Class IX-X Tuition

Find the A.P whose 10th term is 5 and 18th term is 77?

Tapas Bhattacharya

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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 Tuition/Class IX-X Tuition

The sum of the first six terms of an A.p is zero and the fourth term is 2. Find the sum of its first 30 terms?

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 Tuition/Class IX-X Tuition

An A.P consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term?

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 Tuition/Class IX-X Tuition

Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers?

Tapas Bhattacharya

Tutor

Let the 3 terms are (a - b), a, and (a + b). So, (a - b) + a + ( a + b) = 27=> 3a=27 => a=9. Now the product is 648. So, (9 - b) x 9 x (9 + b) = 648=> (81 - b^2) = 648/9 = 72=> b^2 = 81 - 72=9 => b = =3 or b = -3. For, b = 3, the numbers are: (9 - 3), 9, (9 + 3) = 6, 9, 12 or (9 - (-3)), 9... read more
Let the 3 terms are (a - b), a, and (a + b). So, (a - b) + a + ( a + b) = 27=> 3a=27 => a=9. Now the product is 648. So, (9 - b) x 9 x (9 + b) = 648=> (81 - b^2) = 648/9 = 72=> b^2 = 81 - 72=9 => b = =3 or b = -3. For, b = 3, the numbers are: (9 - 3), 9, (9 + 3) = 6, 9, 12 or (9 - (-3)), 9 , (9 + (-3)) = 12, 9, 6 The three numbers are 6, 9, 12 or 12, 9, 6 (Answer) read less
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