Read Symbol defintion
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Q01 |
- What is arithmetic sereis ?
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Q02 |
- Formula of arithmetic series
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Q03 |
- Find sum of series 1 + 2 + 3 + 4 + ..... + n
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Q04 |
- Arithmetic means
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Q05 |
- Sequence of 1, 3, 6, 10, 15, 21, ......
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Q06 |
- Sequence of 1, 4, 9, 16, 25, 36, 49, ....
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Q07 |
- Sequence of 1, 8, 27, 64, 125, ..........
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Q08 |
- Sum of numbers between 200 and 400 which can be divided by 2 and 5
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Q09 |
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Q10 |
- Formula and outlines
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Answers
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Q01. What is arithmetic series ?
- A series a1 + a2 + a3 + ...... + an
- is called an arithmetic seires if, and only if, the difference between consecutive terms
are all equal, that is
- (a2 - a1) = (a3 - a2) = (a4 - a3) = ........
- The difference is call common difference.
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Q02. Formula
The terms
- Let first term be T(1) = a and common difference is d.
- Then 2nd term is T(2) = a + 1*d
- Then 3rd term is T(3) = a + 2*d
- Then 4th term is T(4) = a + 3*d
- Hence the nth term is T(n) = a + (n-1)*d
The sum
- Let the sum of n terms be S(n)
- Then S(n) = Sum[T(n)] = n*(T(1) + T(n))/2.
- or S(n) = n*(2*a + (n-1)*d)/2.
Prove the formula
- S(n) = (a + 0*d) + (a + 1*d) + (a + 2*d) + ..... + (a + (n-1)*d) .... (1)
- S(n) = (a+(n-1)*d) + (a+(n-2)*d) + + (a + 0*d) ........ (2)
- Add the coresponding terms in (1) and (2)
- 2*S(n) = (2*a+(n-1)d) + (2*a+(n-1)*d) + ........ + (2*a+(n-1)*d)
- There are n terms of (2*a + (n-1)*d)
- Hence 2*S(n) = n*(2*a + (n-1)*d)
- and S(n) = n*(2*a + (n-1)*d)/2
Example : Find sum of 1 + 3 + 5 + 7 + ..... + n.
- The common difference is d = 3 and 1st term is a = 1.
- Hence S(n) = n*(2*a + (n-1)*d)/2 = n*(2+2*(n-1))/2 = n^2.
- That is
- T(2) = 1 + 3 = 4 = 2^2
- T(3) = 1 + 3 + 5 = 9 = 3^2
- T(4) = 1 + 3 + 5 + 7 = 16 = 4^2
- Hence T(100) = 100^2 = 10000.
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Q03. Find sum of series 1 + 2 + 3 + 4 + ..... + n
- This is A.P.
- Hence a = 1 and d = 1.
- Use formula we have S(n) = n*(2*a + (n-1)*d)/2 = n*(n+1)/2
- Hence we have Sum(n) = n*(n+1)/2.
Formula
- Sum[1] = n
- Sum[n] = n*(n+1)/2
- Sum[2*n-1] = n^2 where (2*n-1) are odd integers.
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Q04. Arithmatic means
- Definition
- In arithmetic series between the 1st and last term are called arithmetic mean.
- Note : Mid point between two points is arithmetic mean.
- If a,b,c are 3 consecutive terms in an A.P. then arithmetic mean is b = (a+c)/2.
- Since a,b,c are consecutive terms in arithmetic series
- Hence (b-a) = (c-b)
- Hence 2*b = (a + c) or b = (b+c)/2.
Example : x^2, (4*x+4) and (7*x-4) are consecutive terms of an A. P. Find the terms.
- Since (4*x+4) - x^2 = (7*x-4) - (4*x+4)
- Hence -x^2 + (4*x - 7*x + 4*x) + (4 + 4 + 4 ) = 0
- Simplify -x^2 + x + 12 = 0 or x^2 - x - 12 = 0
- Hence (x + 3)*(x - 4) = 0
- Hence x = -3 or x = 4.
- If x = -3, the terms are 9, -8, -25.
- if x = 4 the terms are 16, 20, 24.
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Q05. Sequence of 1, 3, 6, 10, 15, 21, ...
Prove that the first difference of this sequence is arithmetic sequence
- 1st diff : (3-1), (6-3), (10-6), (15-10), (21-15), ....
- 1st diff : 2, 3, 4, 05, 06, 07 ...... with common difference is 1.
- Hence 1st difference of this sequence is arithmetic sequence.
Find next term of this sequnce
- Sequence : 1, 3, 6, 10, 15, 21, x, .....
- 1st diff : 2, 3, 4, 05, 06, 07 ....
- The 7th term x = 21 + 7 = 28.
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Q06. Sequence of 1, 4, 9, 16, 25, 36, 49, ....
Prove that the first difference of this sequence is arithmetic sequence
- Sequence : 1, 4, 9, 16, 25, 36, 49, .....
- 1st diff : (4-1), (9-4), (16-9), (25-16), (36-25), (49-36), ....
- 1st diff : 3, 5, 7, 09, 11, 13, 15, 17, ...... with common difference is 2.
- Hence 1st difference of this sequence is arithmetic sequence.
Find next term of this sequnce
- Sequence : 1, 4, 9, 16, 25, 36, 49, x, .....
- 1st diff : 3, 5, 7, 09, 11, 13, 15, 17, ....
- The 8th term x = 49 + 15 = 64.
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Q07. Sequence of 1, 8, 27, 64, 125, .....
Prove that the 2nd difference of this sequence is arithmetic sequence
- 1st difference
- (8-1), (27-8), (64-27), (125-64), ....
- 7, 19, 37, 61, ....
- 2nd difference
- (19-7), (37-19), (61-37), ....
- 12, 18, 24, .... with common diference is 2.
- Hence 2nd difference is arithmetic sequence.
Find next term
- 01, 08, 27, 64, 125, x, .....
- 07, 19, 37, 61, y
- 12, 18, 24, 30,
- Hence y = 61 + 30 = 91 and x = 125 + y = 125 + 91 = 216 = 6^3
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Q08. Sum of numbers between 200 and 400 which can be divided by 2 and 5
Sum of numbers can be divided by 5
- T(1) = 200 and d = 5
- T(n) = 400 = T(1) + (n-1)*d = 200 + (n-1)*5
- Hence 400 = 200 + 5*(n-1)
- Hence n = 41.
- S(n) = n*(2*a + (n-1)*d)/2 = 41*(2*200 + (41-1)*5)/2 = 12300
Sum of numbers can be divided by 2
- T(1) = 200 and d = 2
- T(n) = 400 = T(1) + (n-1)*2 = 200 + (n-1)*2
- Hence 400 = 200 + 2*(n-1)
- Hence n = 101.
- S(n) = n*(2*a + (n-1)*d)/2 = 101*(2*200 + (101-1)*2)/2 = 30300
Sum of numbers can be divided by 10
- T(1) = 200 and d = 10
- T(n) = 400 = T(1) + (n-1)*10 = 200 + (n-1)*10
- Hence 400 = 200 + 10*(n-1)
- Hence n = 21.
- S(n) = n*(2*a + (n-1)*d)/2 = 21*(2*200 + (21-1)*10)/2 = 6300
The solution
- Since sum of numbers divided by 10 include both case divide by 2 and 5.
- Hence the answer is 12300 + 30300 - 6300 = 37300.
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Q09.
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Q10. Formula and outlines
Formula for A.P.
- The nth term : T(n) = a + (n-1)*d
- The sum of n terms : Sum[T(n)] = n*(2*a + (n-1)*d)/2
- If a,b,c are consecutive A.P. terms, then b = (a + c)/2
Other series
If T(n) = 1, then Sum[1] = n
If T(n) = n, then Sum[n] = n*(n+1)/2
if T(n) = 2*n - 1, then Sum[2*n-1] = n^2
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