Last updated at March 19, 2021 by Teachoo

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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 = 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 need to find d We know that an = a + (n – 1) d Putting a = 7, n = 13 and an = 35 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) Putting n = 13, a = 7, 𝑙 = a13 = 35 = 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 Finding a We know that an = a + (n – 1) d Putting d = 3, n = 12 and a12 = 37 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) Putting n = 12 , a = 4, 𝑙 = a12 = 37 S12 = 12/2 (4+37) = 12/2×41 = 6 ×41 = 246 Ex 5.3, 3 (iv) Given a3 = 15, S10 = 125, find d and a10. For a3 = 15 We know that an = a + (n – 1) d Putting an = a3 = 15 & n = 3 15 = a + 2d 15 – 2d = a a = 15 – 2d For S10 = 125 We know that Sn = 𝑛/2(𝑎+(𝑛−1)𝑑) Putting Sn = S10 = 125, n = 10 S10 = 10/2 (2𝑎+(10−1)𝑑) 125 = 5 (2a + 9d) 125/5=2𝑎+9𝑑 25 = 2a + 9d 25 – 9d = 2a a = (𝟐𝟓 − 𝟗𝒅)/𝟐 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 Putting n = 10 , a = 17 & d = –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) Putting Sn = S9 = 75 , n = 9 , d = 5 75 = 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 = 18a (−210)/18=𝑎 (−35)/3=𝑎 a = (−𝟑𝟓)/𝟑 Now, we need to find a9 We know that an = a + (n – 1) d Putting a = ( 35)/3, d = 5 & n = 9 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 a = 2, d = 8, Sn = 90 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 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 2n + 9 = 0 2n = –9 n = (−𝟗)/𝟐 n – 5 = 0 n = 5 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 a = 2, n = 5, d = 8 a5 = 2 + (5 – 1) 8 a5 = 2 + (4)8 a5 = 2 + 32 a5 = 34 Therefore, n = 5 & a5 = 34 Ex 5.3, 3 In an AP (vii) Given a = 8, an = 62, Sn = 210, find n and d. Given a = 8, an = 62, Sn = 210, Since there are n terms, 𝑙 = an = 62 We use the formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting a = 8, Sn = 210, 𝑙 = an = 62 210 = 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 Putting an = 62, a = 8, n = 6 62 = 8 + (6 – 1) × 𝑑 62 = 8 + 5d 62 – 8 = 5d 54 = 5d 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 Since there are n terms, 𝑙 = an = 4 We use the formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting Sn = −14, 𝑙 = an = 4 –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 Putting an = 4 , d = 2 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 – 10a + 8a– 80 = 0 a (a – 10) + 8 (a – 8) = 0 (a – 10) (a + 8) = 0 So, a = 10 or a = –8 Taking a = 10 n = (− 28)/(𝑎 + 4) n = (− 28)/(10 + 4) n = (− 28)/14 n = –2 Since n is number of terms, it cannot be negative So, n = –2 is not possible ∴ a = 10 is not possible Taking a = – 8 n = (− 28)/(𝑎 + 4) n = (− 28)/(− 8 + 4) n = (− 28)/( − 4) n = 7 So, n = 7 So, answer is n = 7 and 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 a = 3, n = 8 , Sn = 192 192 = 8/2(2×3+(8−1)×𝑑) 192 = 4 (6 + 7 d) 192/4 = 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) Putting Sn = 144, n = 9 & 𝑙 = 28 144 = 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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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.