Ex 5.2, 3 (ii) - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at Aug. 3, 2021 by Teachoo
Transcript
Ex 5.2, 3 In the following APs find the missing term in the boxes (ii) ⎕, 13, ⎕, 3 Given a2 = 13 & a4 = 3 We know that an = a + (n – 1) d Given 2nd term is 13 a2 = a + (2 – 1) d 13 = a + d 13 – d = a a = 13 – d Given 4th term is 3 a4 = a + (4 – 1) d 3 = a + 3d 3 – 3d = a a = 3 – 3d From (1) & (2) 13 – d = 3 – 3d 13 – 3 = d – 3d 10 = –2d 10/(−2) = d – 5 = d d = – 5 Putting value of d in (1) a = 13 – d = 13 – ( – 5) = 13 + 5 = 18 Now, We need to find a3 a3 = a + (n – 1) d = 18 + (3 – 1) × – 5 = 18 + 2 × – 5 = 18 – 10 = 8 So, a3 = 8 Hence, , 13, , 3
Ex 5.2
Ex 5.2, 1
Ex 5.2, 2 (i) (MCQ)
Ex 5.2, 2 (ii) (MCQ) Important
Ex 5.2, 3 (i)
Ex 5.2, 3 (ii) You are here
Ex 5.2, 3 (iii) Important
Ex 5.2, 3 (iv)
Ex 5.2, 3 (v) Important
Ex 5.2, 4
Ex 5.2, 5 (i)
Ex 5.2, 5 (ii) Important
Ex 5.2, 6 Important
Ex 5.2, 7
Ex 5.2, 8
Ex 5.2, 9
Ex 5.2, 10
Ex 5.2, 11
Ex 5.2, 12 Important
Ex 5.2, 13 Important
Ex 5.2, 14 Important
Ex 5.2, 15
Ex 5.2, 16
Ex 5.2, 17 Important
Ex 5.2, 18
Ex 5.2, 19
Ex 5.2, 20 Important
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
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 10 years. He provides courses for Maths and Science at Teachoo.