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Ex 5.3, 1 - Find sum of APs. (i) 2, 7, 12,... to 10 terms - Finding sum of n terms

  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
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Ex 5.3 ,1 Find the sum of the following APs. (i) 2, 7, 12 ,…., to 10 terms. 2, 7, 12,….,to 10 term We know that Sum of AP = 𝑛/2 (2a + (n – 1) d) Here n = 10, a = 2, & d = 7 – 2 = 2 Putting these in formula , Sum = 𝑛/2 (2a + (n – 1) d) = 10/2 (2 × 2 + (10 – 1) × 5) = 5(3 + 9 × 5) = 5 (4 + 45) = 5 (49) = 245 Ex 5.3 ,1 Find the sum of the following APs. (ii) − 37, − 33, − 29 ,…, to 12 terms – 37, – 33, – 29 to 12 terms We know that Sum = 𝑛/2 (2a + (n – 1)d) Here n = 12, a = – 37 & d = – 33 – ( – 37) = – 33 + 37 = 4 Putting values in formula Sum = 𝑛/2 (2a + (n – 1) d) = 12/2 (2 × (–37) + (12 – 1) × 4) = 6 ( –74 + 11 × 4) = 6 ( –74 + 44) = 6 ( –30) = – 180 Ex 5.3 ,1 Find the sum of the following APs. (iii) 0.6, 1.7, 2.8 ,…….., to 100 terms 0.6, 1.7, 2.8 to 100 terms We know that Sum = 𝑛/2 (2a + (n – 1) d) Here n = 100, a = 0.6, & d = 1.7 – 0.6 = 1.1 Putting in formula Sum = 𝑛/2 (2a + (n – 1) d) = 100/2 (2 × 0.6 + (100 – 1) × 1.1) = 50(1.2 + 99 × 1.1) = 50(1.2 + 108.9) = 50 (110.1) = 50 × 110 . 1 = 5505 Ex 5.3 ,1 Find the sum of the following APs. (iv)  1/15, 1/12, 1/10 ,………, to 11 terms 1/15 , 1/12 , 1/10 , ….. to 11 terms We know that Sum = 𝑛/2 (2a + (n – 1) d) Here n = 11 , a = 1/15 & d = 1/12 – 1/15 = (15 − 12)/(12 × 15) = 3/180 = 1/60 Putting values in formula Sum = 𝑛/2 (2a + (n – 1) d) = 11/2 (2 × 1/15 + (11 – 1) × 1/60 ) = 11/2 (2/15 + 10 × 1/60 ) = 11/2 (2/15 + 10/60 ) = 11/2 ((2 × 4 + 10)/60) = 11/2 ((8 + 10)/60) = 11/2 × 18/60 = 11× 9/60 = 11 × 3/20 = 33/20 = 1.65

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