Check sibling questions

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

This video is only available for Teachoo black users


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

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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.