Ex 5.3

Chapter 5 Class 10 Arithmetic Progressions
Serial order wise

### Transcript

Ex 5.3, 3 In an AP (viii) Given an = 4, d = 2, Sn = −14, find n and a. Given an = 4, d = 2, Sn = –14 Since there are n terms, 𝑙 = an = 4 We use the formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting Sn = −14, 𝑙 = an = 4 –14 = 𝑛/2 (𝑎+4) –14 × 2 =𝑛(𝑎+4) –28 = n (a + 4) (−28)/(𝑎 + 4)=𝑛 n = (−𝟐𝟖)/(𝒂 + 𝟒) Also we know that an = a + (n – 1) d Putting an = 4 , d = 2 4 = a + (n – 1) × 2 4 = a + 2n – 2 4 + 2 = a + 2n 6 = a + 2n Putting n = (− 𝟐𝟖)/(𝒂 + 𝟒) 6 = a + 2((− 28)/(𝑎 + 4)) 6 = a − 56/(𝑎 + 4) 6 = (𝑎(𝑎 + 4) − 56)/(𝑎 + 4) 6 = (𝑎2 + 4𝑎 − 56)/(𝑎 + 4) 6 (a + 4) = a2 + 4a – 56 6a + 24 = a2 4a – 56 0 = a2 + 4a – 56 – 6a – 24 0 = a2 – 2a – 80 a2 – 2a – 80 = 0 a2 – 10a + 8a– 80 = 0 a (a – 10) + 8 (a – 8) = 0 (a – 10) (a + 8) = 0 So, a = 10 or a = –8 Taking a = 10 n = (− 28)/(𝑎 + 4) n = (− 28)/(10 + 4) n = (− 28)/14 n = –2 Since n is number of terms, it cannot be negative So, n = –2 is not possible ∴ a = 10 is not possible Taking a = – 8 n = (− 28)/(𝑎 + 4) n = (− 28)/(− 8 + 4) n = (− 28)/( − 4) n = 7 So, n = 7

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.