# Example 19 - Chapter 9 Class 11 Sequences and Series

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 19, Find the sum to n terms of the series: 5 + 11 + 19 + 29 + 41… It is not an AP or a GP Let Sn = 5 + 11 + 19 + 29 + 41 ... + an–1 + an Sn = 0 + 5 + 11 + 19 + 41 ... + an–2 + an–1 + an Subtracting (2) from (1) Sn – Sn = 5 – 0 + [(11 – 5) + (19 – 11) + (29 – 19) + ...(an–1 – an–2 ) + (an – an – 1)] – an 0 = 5 + [6 + 8 + 10 + 12 + ...an–1 ] – an an = 5 + [6 + 8 + 10 + 12 + ... + (n – 1) terms] 6 + 8 + 10 + 12 + ... + (n – 1) term is an AP With first term a = 6 & common difference = d = 8 – 6 = 2 Sum of n terms of an AP = 𝑛/2 (2a + (n – 1)d) Putting n = n – 1 & a = 6 & d = 2 [6 + 8 + 10 + 12 + ... + (n – 1) terms] = (n−1)/2 [ 2(6) +((n – 1) – 1)2 ] = (n−1)/2 [ 12+(n – 1 – 1)2 ] = (n−1)/2 [ 12+(n – 2)2 ] = (n−1)/2 [12 + 2n – 4] = (n−1)/2 [8 + 2n] = (n−1)/2 × 2[4 + n] = (n – 1) (n + 4) Thus, [6 + 8 + 10+ … upto (n –1) terms] = (n – 1) (n + 4) Now, an = 5 + [6 + 8 + 10 + 12 + ... + (n – 1) terms] Putting values an = 5 + (n – 1)(n – 4) an = 5 + n(n + 4) – 1(n + 4) an = 5 + (n2 + 4n) – n – 4 an = 5 + n2 + 4n – n – 4 an = n2 + 3n + 1 Now = (n(n + 1)(2n + 1))/6 + 3((n(n + 1))/2) + n = (𝑛(n + 1)(2n + 1))/6 + 3/2 n(n + 1) + 𝑛/1 = (𝑛(n + 1)(2n + 1) + 9(n + 1)) + 6n)/6 = n(((n + 1)(2n + 1) + 9(n + 1) + 6)/6) = n((𝑛(2n + 1) + 1(2n + 1) + 9n + 9 + 6)/6) = n ((2n2 + 2n + n + 1 + 9n + 9 + 6)/6) = n((2n2 + 12n + 16)/6) = n((2(n2 +6n +8))/6) = n/3 (n2 + 6n +8) = n/3 [n(n + 4) + 2(n + 4)] = n/3 [(n + 2)(n + 4)] = (n(n + 2)(n + 4) )/3 Thus, the required sum is (n(n + 2)(n + 4) )/3

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6 Important

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19 Important You are here

Example 20

Example 21 Important

Example 22

Example 23 Important

Example 24 Important

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.