Ex 5.2, 3

Transcript

Ex 5.2 ,3 (Method 1) In the following APs find the missing term in the boxes (i) 2, ,26 Let 2nd term = x So, AP is 2, x, 26 Since it is an AP, common difference is same ∴ Second term – First term = Third term – Second term x – 2 = 26 – x x + x = 2 + 26 2x = 28 x = 28/2 x = 14 So, 2, ,26 Ex 5.2 ,3 (Method 2) In the following APs find the missing term in the boxes (i) 2, ,26 Given a = 2 a3 = 26 We need to find a2 We know that an = a + (n – 1) d Here , a3 = 26 , a = 2 ,n = 3 First we find d Putting values in formula an = a + (n – 1) d 26 = 2 + (3 – 1)d 26 = 2 + 2d 26 – 2 = 2d 24 = 2d 24/2 = d 12 = d d = 12 Now, Second term = a2 = a + (n – 1) d = 2 + (2 – 1) × 12 = 2 + 1 × 12 = 2 + 12 = 14 So, 2, ,26 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 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 ,3 In the following APs find the missing term in the boxes (iii) 5, , , 91/2 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 So, 5, , , 91/2 Ex 5.2 ,3 In the following APs find the missing term in the boxes (iv) – 4, , , , , 6 Given a = –4, a6 = 6 We know that an = a + (n – 1) d Putting values a6 = – 4 + (6 – 1) × d 6 = – 4 + 5 d 6 + 4 = 5d 10 = 5d 10/5 = d 2 = d d = 2 Now , a = – 4, d = 2 We can find a2, a3, a4, a5 a2 = a + (2 – 1) d = a + d = – 4 + 2 = – 2 a3 = a + (3 – 1) d = a + 2d = – 4 + 2 × 2 = – 4 + 4 = 0 a4 = a + (4 – 1) d = a + 3 d = – 4 + 3 × 2 = 4 + 6 = 2 a5 = a + (5 – 1) d = a + 4d = – 4 + 4 × 2 = – 4 + 8 = 4 Hence, – 4, , , , , 6 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 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 So, a3 = a + (3 – 1) d = a + 2d = 53 + 2 × (–15) = 53 – 30 = 23 Also, a4 = a + (4 – 1) d = a + 3 d = 53 + 3 × (–15) = 53 – 45 = 8 a5 = a + (5 – 1) d = a + 4d = 53 + 4 × (–15) = 53 – 60 = –7 Hence, ,38, , , , - 22