# CLASS-10ARITHMETIC PROGRESSION - FINDING SUM OF AN A.P.

Sum of e-Terms of an A.P.-

Theorem 3.) Prove that the sum of ‘e’ terms of an A.P. in which the first term is ‘a’, the common difference is ‘d’ and the last term is ‘l’, is given by –

e                         e

Sₑ = ------ (a + l) and Sₑ = ------ {2a + (e – 1) d}

2                         2

Proof – Consider an A.P. having ‘e’ terms in which, first term = a, common difference = d, and last term = l.

Then, l = a + (e – 1) d ……………(i)      [where l = eth term]

We may write the given A.P. as

a, (a + d), (a + 2d), (a + 3d),………………,(l – 2d), (l – d), l

let, Sₑ be the sum of ‘e’ terms of this A.P. Then

Sₑ = a + (a + d) + (a + 2d) + ……………+ (l – 2d) + (l – d) + l …………….(ii)

Writing the above series in the reverse order, we get –

Sₑ = l + (l – d) + (l – 2d) + ………….+ (a + 2d) + (a + d) + a …………………(iii)

Adding the corresponding terms of (ii) and (iii), we get,

2Sₑ = (a + l) + (a + l) + (a + l) + …………….. e times

=> 2Sₑ = e (a + l)

e

=> Sₑ = ------- (a + l)

2

e

=> Sₑ = ------- [a + {a + (e – 1) d}]              [using (i)]

2

e

=> Sₑ = ------- {2a + (e – 1) d}

2

e                         e

Hence, Sₑ = ------ (a + l) and Sₑ = ------ {2a + (e – 1) d}

2                         2

Example.1) Find the sum of first 15 terms of the A.P. 3, 8, 13, 18, 23,………………

Ans.) The given A.P. is 3, 8, 13, 23,………………………

Here, a = 3, d = (8 – 3) = (13 – 8) = (18 – 13) = (23 – 18) = 5 (constant), and e = 15

e

Using the formula, Sₑ = ------- [2a + (e – 1) d], we get -

2

15

S₁₅ = ------- [(2 X 3) + {(15 – 1) X 5}]

2

15

= ------ {6 + (14 X 5)}

2

15 X 76

= ----------- = 570

2

Hence, the sum of the first 15 terms of the given A.P. is 570     (Ans.)