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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 March 22, 2023 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