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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) 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
Last updated at Aug. 3, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
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