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Ex 5.3, 10 (ii) - Chapter 5 Class 10 Arithmetic Progressions
Last updated at May 29, 2023 by Teachoo
Ex 5.3
Ex 5.3, 1 (i)
Ex 5.3, 1 (ii)
Ex 5.3, 1 (iii) Important
Ex 5.3, 1 (iv)
Ex 5.3, 2 (i)
Ex 5.3, 2 (ii)
Ex 5.3, 2 (iii) Important
Ex 5.3, 3 (i)
Ex 5.3, 3 (ii)
Ex 5.3, 3 (iii)
Ex 5.3, 3 (iv) Important
Ex 5.3, 3 (v)
Ex 5.3, 3 (vi) Important
Ex 5.3, 3 (vii)
Ex 5.3, 3 (viii) Important
Ex 5.3, 3 (ix)
Ex 5.3, 3 (x)
Ex 5.3, 4
Ex 5.3, 5
Ex 5.3, 6 Important
Ex 5.3, 7
Ex 5.3, 8
Ex 5.3, 9
Ex 5.3, 10 (i)
Ex 5.3, 10 (ii) Important You are here
Ex 5.3, 11 Important
Ex 5.3, 12
Ex 5.3, 13
Ex 5.3, 14 Important
Ex 5.3, 15
Ex 5.3, 16 Important
Ex 5.3, 17
Ex 5.3, 18 Important
Ex 5.3, 19 Important
Ex 5.3, 20 Important
Chapter 5 Class 10 Arithmetic Progressions
Serial order wise
Transcript
Ex 5.3, 10 Show that a1, a2 … , an , … form an AP where an is defined as below (ii) an = 9 − 5n . Also find the sum of first 15 terms in each case Since an = 9 – 5n Hence, the series is 4, –1, –6, …… Since difference is same, it is an AP Common difference = d = –1 – 4 = –5 Taking n = 1 a1 = 9 – 5 × 1 a = 9 – 5 a = 4 Taking n = 2 a2 = 9 – 5 × 2 a2 = 9 – 10 a2 = –1 Taking n = 3, a3 = 9 – 5 × 3 a3 = 9 – 15 a3 = –6 Now we have to find sum of first 15 terms i.e. S15 We know that Sn = 𝒏/𝟐 (𝟐𝒂+(𝒏−𝟏)𝒅) Putting n = 15, a = 4, d = −5 S15 = 15/2 (2 × 4+(15−1)×−5) S15 = 15/2 (8+14×−5) S15 = 15/2 (8−70) S15 = 15/2× − 62 S15 = –465
