Ex 5.2
Ex 5.2, 2 (i) (MCQ)
Ex 5.2, 2 (ii) (MCQ) Important
Ex 5.2, 3 (i)
Ex 5.2, 3 (ii)
Ex 5.2, 3 (iii) Important
Ex 5.2, 3 (iv)
Ex 5.2, 3 (v) Important You are here
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
Last updated at April 16, 2024 by Teachoo
Ex 5.2, 3 In the following APs find the missing term in the boxes (v) ⎕, 38, ⎕ , ⎕, ⎕, − 22 Given a2 = 38 & a6 = – 22 We know that an = a + (n – 1) d Given 2nd term is 38 a2 = a + (2 – 1) d 38 = a + d 38 – d = a a = 38 – d Given 6th term is –22 a4 = a + (6 – 1) d –22 = a + 5d –22 – 5d = a a = –22 – 5d From (1) & (2) 38 – d = –22 – 5d 38 + 22 = d – 5d 60 = –4d 60/(−4) = d –15 = d d = –15 Putting value of d in (1) a = 38 – d = 38 – ( – 15) = 38 + 15 = 53 Hence, a = 53, d = –15 We need to find a3, a4, a5 We know that an = a + (n – 1)d a3 a3 = a + (3 – 1) d = a + 2d = 53 + 2(−15) = 53 − 30 = 23 a4 a4 = a + (4 – 1) d = a + 3d = 53 + 3(−15) = 53 − 45 = 8 a5 a3 = a + (5 – 1)d = a + 4d = 53 + 4(−15) = 53 − 60 = −7 Hence, ,38, , , , - 22