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Ex 5.2, 3 (iii) - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at Aug. 3, 2021 by Teachoo
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)
Ex 5.2, 3 (iii) Important You are here
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
Serial order wise
Transcript
Ex 5.2, 3 In the following APs find the missing term in the boxes (iii) 5, ⎕, ⎕, 9 1/2 Given a = 5 , a4 = 91/2 = (2(9) + 1)/2 = 19/2 We know that an = a + (n – 1) d Putting values a4 = a + (4 – 1) d 19/2 = 5 + 3d 19/2 – 5 = 3d (19 − 2(5))/2 = 3d 9/2 = 3d 9/2 × 1/3 = d 3/2 = d d = 𝟑/𝟐 Hence, a = 5 and d = 3/2 Now, We need to find a2 & a3 Hence, a2 = 13/2 = 61/2 a3 = 8 a2 an = a + (n – 1) d a2 = a + (2 – 1)d = a + d = 5 + 3/2 = (5(2) + 3)/2 = 𝟏𝟑/𝟐 a3 an = a + (n – 1) d a3 = a + (3 – 1)d = a + 2d = 5 + 2(3/2) = 5 + 3 = 8 So, 5, , , 91/2
