# CLASS-10ARITHMETIC PROGRESSION - ARITHMETIC SERIES

ARITMETIC SERIES -

By adding the terms of an A.P., we get the corresponding arithmetic series.

Example.) On adding the terms of the A.P. 6, 2, -2, -6, -10, ………….,

We get the arithmetic series [6 + 2 + (-2) + (-6) + (-10) + ………………]

To Find the General Term of an A.P.

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

Proof :- In the given A.P. we have first term = a, and common difference = d.

So, the given A.P. may be written as –

a, (a + d), (a + 2d), (a + 3d), (a + 4d),……………….

In this A.P. we have –

First term, T₁ = a = a + (1 – 1) d,

Second term, T₂ = a + d = a + (2 – 1) d,

Third term, T₃ = a + 2d = a + (3 – 1) d, and so on

………………. …………………. ………………….. ……………………

………………. …………………. ………………….. ……………………

eth term, Tₑ = a + (e – 1) d

for your information, the eth term of an A.P. is called its general terms.

Example.1) Show that the numbers 5, 10, 15, 20, 25, ................. form an A.P. Find its –

(i) eth term (ii) 15th term

Ans.) The given progression is 5, 10, 15, 20, 25, .................

We have, (10 – 5) = 5

(15 – 10) = 5

(20 – 15) = 5

(25 – 20) = 5

So, we find that, (10 – 5) = (15 – 10) = (20 – 15) = (25 – 20) = 5 (constant)

So, the given number from an A.P.

Its first term is a = 5, and the common difference d = 5

(i) its eth term is given by

Tₑ = a + (e – 1) d

= 5 + (e – 1) 5 = 5 + 5e – 5

= 5e .................(i)           (Ans.)

Now, when e = 15, then 15th term of given A.P. is –

Tₑ = a + (e – 1) d

T₁₅ = 5 + (15 – 1) 5 [putting e = 15]

= 5 + (14 X 5)

= 5 + 70

= 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

Ans.) The given progression is 17, 11, 5, -1, -7, - 13,................

We have, (11 – 17) = - 6

(5 – 11) = - 6

(-1 – 5) = - 6

[-7 – (-1)] = (-7 + 1) = - 6

[- 13 – (- 7)] = (- 13 + 7) = - 6

So, we have, (11 – 17) = (5 – 11) = (-1 – 5) = [-7 – (-1)] = [- 13 – (- 7)] = - 6 (constant)

So, the given numbers form an A.P.

Its first term, a = 17, and common difference, d = - 6

(i) Its eth term is given by

Tₑ = a + (e – 1) d

=> Tₑ = 17 + (e – 1) (- 6)

= 17 + (6 – 6e)

= 23 – 6e

So, eth term, Tₑ = (- 6e + 23)       (Ans.)

(ii) Putting e = 10 in A.P. we get

Tₑ = (- 6e + 23)

T₁₀ = [{(- 6) X 10} + 23]

= (- 60 + 23) = - 37            (Ans.)

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

Ans.) It is given that, Tₑ = (5e – 6)

Now, putting e = 1, we get T₁ = [(5 X 1) – 6] = (5 – 6) = - 1

e = 2, we get T₂ = [(5 X 2) – 6] = (10 – 6) = 4

e = 3, we get T₃ = [(5 X 3) – 6] = (15 – 6) = 9

and, so on ...........................

so, its 1st term is T₁ = - 1 ......................(i)      (Ans.)

Its common difference, d = (T₂ - T₁) = {4 – (- 1)} = (4 + 1) = 5 .................(ii)          (Ans.)

Putting e = 20, and we get Tₑ = (5e – 6)

T₂₀ = {(5 X 20) – 6} = (100 – 6) = 94 ...................(iii)      (Ans.)

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