# Ex 5.3, 3

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 5.3 ,3 In an AP (i) Given a = 5, d = 3, an = 50, find n and Sn. Given a = 5 , d = 3 , an = 50 We know that an = a + (n – 1) d Putting values 50 = 5 + (n – 1) ×3 50 = 5 + 3n – 3 50 = 2 + 3n 50 – 2 = 3n 48 = 3n 48/3=𝑛 n = 16 Now we need to find Sn Sn = 𝑛/2(2𝑎+(𝑛−1)𝑑 Putting n = 16, a = 5, d = 3 Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) = 16/2 (2×5+(16−1)×3) = 8 (10+15×3) = 8(10+45) = 8 ×55 = 440 Ex 5.3 ,3 In an AP (ii) Given a = 7, a13 = 35, find d and S13. Given a = 7, a13 = 35 We know that an = a + (n – 1) d Here, a13 = 35, n = 13, a = 7 We need to find d Putting values in formula an = a + (n – 1) d 35 = 7 + (13 – 1) ×𝑑 35 = 7 + 12d 35 – 7 = 12d 28 = 12 d 28/12=𝑑 7/3=𝑑 d = 7/3 Now we need to find S13 We can use formula Sn = 𝑛/2 (𝑎+𝑙) Here, n = 13, a = 7, 𝑙 = a13 = 35 Putting values in formula Sn = 𝑛/2 (𝑎+𝑙) = 13/2(7+35) = 13/2×42 = 13 ×21 = 273 Ex 5.3 ,3 In an AP (iii) Given a12 = 37, d = 3, find a and S12. Given a12 = 37, d = 3 We know that an = a + (n – 1) d Here , an = a12 = 37 , n = 12 , d = 3 We can find a Putting values an = a + (n – 1) d 37 = a + (12 – 1) ×3 37 = a + 11 ×3 37 = a + 33 37 – 33 = a 4 = a a = 4 Now we can find (s12) by using formula Sn = 𝑛/2 (𝑎+𝑙) Here n = 12 , a = 4, 𝑙 = a12 = 37 Putting values S12 = 12/2 (4+37) = 12/2×41 = 6 ×41 = 246 Ex 5.3 ,3 In an AP (iv) Given a3 = 15, S10 = 125, find d and a10. From (1) and (2) 15 – 2d = (23−9𝑑)/2 30 – 4d = 25 – 9d 5 = – 5d 5/(− 5)=𝑑 d = – 1 Putting value of d in (1) a = 15 – 2d a = 15 – 2 ×(−1) a = 15 + 2 a = 17 Now we need to find a10 an = a + (n – 1) d Here n = 10 , a = 17 & d = – 1 Putting values an = a + (n – 1) d a10 = 17 + (10 – 1) ×−1 a10 = 17 + (10 – 1) ×(−1) a10 = 17 + 9 ×(−1) a10 = 17 – 9 a10 = 8 Ex 5.3 ,3 In an AP (v) Given d = 5, S9 = 75, find a and a9. Given d = 5 , S9 = 75 We use formula Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Here , Sn = S9 = 75 , n = 9 , d= 5 Putting values Sn = 9/2 (2𝑎+(9−1)×5) 75 = 9/2 (2𝑎+8×5) 75 = 9/2(2𝑎+40) (75 × 2)/9=2𝑎+40 150/9=2𝑎+40 150 = 9 (2a + 40) 150 = 9 (2a) + 9(40) 150 = 18a + 360 150 – 360 = 18a – 210 = 18 a (− 210)/18=𝑎 (−35)/3=𝑎 a = (−35)/3 Now, we need to find a9 We know that an = a + (n – 1) d Here a = (− 35)/3 , d = 5 & n = 9 Putting values an = a + (n – 1) d a9 = (− 35)/3+(9−1)×5 a9 = (− 35)/3+8×5 a9 = (− 35)/3+40 a9 = (− 35 + 40 × 3)/3 a9 = (− 35 +120)/3 a9 = 85/3 Ex 5.3 ,3 In an AP (vi) Given a = 2, d = 8, Sn = 90, find n and an. Given a = 2, d = 8, Sn = 90 We can use formula Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Putting values Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) 90 = 𝑛/2 (2×2+(𝑛−1)×8) 90 = 𝑛/2 (4+8𝑛−8) 90 ×2=𝑛(4+8𝑛−8) 180 = n (8n – 4) 180 = 8n2 – 4n – 180 + 8n2 – 4n = 0 – 180 + 8n2 – 4n = 0 8n2 – 4n – 180 = 0 8n2 – 4n – 180 = 0 4 (2n2 – n – 45) = 0 2n2 – n – 45 = 0 2n2 – 10n + 9n – 45 = 0 2n(n – 5) + 9(n – 5) = 0 (2n + 9) (n – 5) = 0 Therefore, n = 5 & n = −9/2 But n cannot be in fraction, So, n = 5 We need to find an, i.e. a5 We know that an = a + (n – 1)d Putting values a5 = 2 + (5 – 1) 8 a5 = 2 + (4)8 a5 = 2 + 32 Ex 5.3 ,3 In an AP (vii) Given a = 8, an = 62, Sn = 210, find n and d. We know that Sn = 𝑛/2 (𝑎+𝑙) Here, a = 8, Sn = 210 Since there are n terms , 𝑙 = an = 62 Putting values in formula Sn = 𝑛/2 (𝑎+𝑙) 210 = 𝑛/2 (8+62) 210 ×2=𝑛×(70) 420 = n ×70 420/70=𝑛 6 = n n = 6 Now we need to find d We can use formula an = a + (n – 1) d Here , an = 62, a = 8, n = 6 Putting values an = a + (n – 1) d 62 = 8 + (6 – 1) ×𝑑 62 = 8 + 5d 62 – 8 = 5d 54 = 5d 54/5=𝑑 d = 54/5 Ex 5.3 ,3 In an AP (viii) Given an = 4, d = 2, Sn = − 14, find n and a. Given an = 4 ,d = 2 ,Sn = – 14 We know that Sn = 𝑛/2 (𝑎+𝑙) Here, d = 2, Sn = –14 Since there are n terms , 𝑙 = an = 4 Putting values – 14 = 𝑛/2 (𝑎+4) – 14 ×2=𝑛(𝑎+4) – 28 = n (a + 4) (− 28)/(𝑎 + 4)=𝑛 n = (− 28)/(𝑎 + 4) Also we know that an = a + (n – 1) d Here, an = 4 , d = 2 Putting values 4 = a + (n – 1) ×2 4 = a + 2n – 2 4 + 2 = a + 2n 6 = a + 2n Putting n = (− 28)/(𝑎 + 4) 6 = a + 2((− 28)/(𝑎 + 4)) 6 = a − 56/(𝑎 + 4) 6 = (𝑎(𝑎 + 4) − 56)/(𝑎 + 4) 6 = (𝑎2 + 4𝑎 − 56)/(𝑎 + 4) 6 (a + 4) = a2 + 4a – 56 6a + 24 = a2 4a – 56 0 = a2 + 4a – 56 – 6a – 24 0 = a2 – 2a – 80 a2 – 2a – 80 = 0 a2 – 10 a + 8a– 80 = 0 a (a – 10) + 8 (a – 8) = 0 (a – 10) (a + 8) = 0 So, a = 10 or a = – 8 Ex 5.3 ,3 In an AP (ix) Given a = 3, n = 8, S = 192, find d. Given a = 3 ,n = 8 , Sn = 192 We know that Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Putting values 192 = 8/2(2×3+(8−1)×𝑑) 192 = 4 (6 + 7 d) 192/4 = 6 + 6 + 7d 48 = 6 + 7d 48 – 6 = 7d 42 = 7d 42/7=𝑑 6 = d d = 6 Ex 5.3 ,3 In an AP (x) Given 𝑙 = 28, S = 144 and there are total 9 terms. Find a. Given 𝑙 = 28 , S = 144 , n = 9 We can use formula Sn = 𝑛/2 (𝑎+𝑙) Here Sn = 144, n = 9 & 𝑙 = 28 We need to find a Putting values Sn = 𝑛/2 (𝑎+𝑙) 144 = 9/2(𝑎+28) (144 × 2)/9=𝑎+28 32 =𝑎+28 32 – 28 = a a = 4

Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .