Check sibling questions

Ex 5.3, 3 (vi) - Chapter 5 Class 10 Arithmetic Progressions

Last updated at May 29, 2023 by

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 13

Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 14
Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 15 Ex 5.3, 3 - Chapter 5 Class 10 Arithmetic Progressions - Part 16

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Book a free demo

Transcript

Ex 5.3, 3 In an AP (vi) Given a = 2, d = 8, Sn = 90, find n and an. Given a = 2, d = 8, Sn = 90 We can use formula Sn = 𝑛/2 (2π‘Ž+(π‘›βˆ’1)𝑑) Putting a = 2, d = 8, Sn = 90 90 = 𝑛/2 (2 Γ— 2+(π‘›βˆ’1) Γ— 8) 90 = 𝑛/2 (4+8π‘›βˆ’8) 90 Γ— 2 =𝑛(4+8π‘›βˆ’8) 180 = n (8n – 4) 180 = 8n2 – 4n –180 + 8n2 – 4n = 0 8n2 – 4n – 180 = 0 8n2 – 4n – 180 = 0 4 (2n2 – n – 45) = 0 2n2 – n – 45 = 0 2n2 – 10n + 9n – 45 = 0 2n(n – 5) + 9(n – 5) = 0 (2n + 9) (n βˆ’ 5) = 0 2n + 9 = 0 2n = –9 n = (βˆ’πŸ—)/𝟐 n – 5 = 0 n = 5 Therefore, n = 5 & n = (βˆ’9)/2 But n cannot be in fraction, So, n = 5 We need to find an, i.e. a5 We know that an = a + (n – 1)d Putting a = 2, n = 5, d = 8 a5 = 2 + (5 – 1) 8 a5 = 2 + (4)8 a5 = 2 + 32 a5 = 34 Therefore, n = 5 & a5 = 34

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.