LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

ARITHMETIC PROGRESSION - ARITHMETIC SERIES

**ARITMETIC SERIES -
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**By adding the terms of an A.P., we get the corresponding arithmetic series.
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**Example.) On adding the terms of the A.P. 6, 2, -2, -6, -10, ………….,
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**We get the arithmetic series [6 + 2 + (-2) + (-6) + (-10) + ………………]
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**To Find the General Term of an A.P.
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__Theorem 1.)__ In an A.P. it is given that its first term is ‘a’ and the common difference is ‘d’. Prove that its eth term is given by

** => Tₑ = a + (e – 1) d
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__Proof :-__ In the given A.P. we have first term = a, and common difference = d.

**So, the given A.P. may be written as –
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**a, (a + d), (a + 2d), (a + 3d), (a + 4d),……………….
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**In this A.P. we have –
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**First term, T₁ = a = a + (1 – 1) d,
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**Second term, T₂ = a + d = a + (2 – 1) d,
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**Third term, T₃ = a + 2d = a + (3 – 1) d, and so on
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**………………. …………………. ………………….. ……………………
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**………………. …………………. ………………….. ……………………
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**eth term, Tₑ = a + (e – 1) d
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**for your information, the eth term of an A.P. is called its general terms.
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**Example.1) Show that the numbers 5, 10, 15, 20, 25, ................. form an A.P. Find its –
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**(i) eth term (ii) 15th term
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**Ans.) The given progression is 5, 10, 15, 20, 25, .................
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**We have, (10 – 5) = 5
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** (15 – 10) = 5
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** (20 – 15) = 5
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** (25 – 20) = 5
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**So, we find that, (10 – 5) = (15 – 10) = (20 – 15) = (25 – 20) = 5 (constant)
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**So, the given number from an A.P.
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**Its first term is a = 5, and the common difference d = 5
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**(i) its eth term is given by
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** Tₑ = a + (e – 1) d
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** = 5 + (e – 1) 5 = 5 + 5e – 5
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** = 5e .................(i) (Ans.)
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**Now, when e = 15, then 15th term of given A.P. is –
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** Tₑ = a + (e – 1) d
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** T₁₅ = 5 + (15 – 1) 5 [putting e = 15]
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** = 5 + (14 X 5)
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** = 5 + 70
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** = 75 ............................(ii) (Ans.)**

**Example.2) Show that the numbers 17, 11, 5, -1, -7, - 13, ................form an A.P. Find its (i) eth term, (ii) 10th term
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**Ans.) The given progression is 17, 11, 5, -1, -7, - 13,................
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**We have, (11 – 17) = - 6
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** (5 – 11) = - 6
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** (-1 – 5) = - 6
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** [-7 – (-1)] = (-7 + 1) = - 6
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** [- 13 – (- 7)] = (- 13 + 7) = - 6
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** So, we have, (11 – 17) = (5 – 11) = (-1 – 5) = [-7 – (-1)] = [- 13 – (- 7)] = - 6 (constant)
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**So, the given numbers form an A.P.
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**Its first term, a = 17, and common difference, d = - 6
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**(i) Its eth term is given by
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** Tₑ = a + (e – 1) d
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** => Tₑ = 17 + (e – 1) (- 6)
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** = 17 + (6 – 6e)
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** = 23 – 6e
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**So, eth term, Tₑ = (- 6e + 23) (Ans.)
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**(ii) Putting e = 10 in A.P. we get
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** Tₑ = (- 6e + 23)
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** T₁₀ = [{(- 6) X 10} + 23]
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** = (- 60 + 23) = - 37 (Ans.)
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**Example.3) The eth term of an A.P. is given by Tₑ = (5e – 6). Find its (i) 1st term, (ii)common difference, and (iii) 20th term
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**Ans.) It is given that, Tₑ = (5e – 6)
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**Now, putting e = 1, we get T₁ = [(5 X 1) – 6] = (5 – 6) = - 1
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** e = 2, we get T₂ = [(5 X 2) – 6] = (10 – 6) = 4
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** e = 3, we get T₃ = [(5 X 3) – 6] = (15 – 6) = 9
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** and, so on ...........................
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**so, its 1st term is T₁ = - 1 ......................(i) (Ans.)
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**Its common difference, d = (T₂ - T₁) = {4 – (- 1)} = (4 + 1) = 5 .................(ii) (Ans.)
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**Putting e = 20, and we get Tₑ = (5e – 6)
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** T₂₀ = {(5 X 20) – 6} = (100 – 6) = 94 ...................(iii) (Ans.)**

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