Ex 5.3, 8 - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at March 22, 2021 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 You are here
Ex 5.3, 9
Ex 5.3, 10 (i)
Ex 5.3, 10 (ii) Important
Ex 5.3, 11 Important
Ex 5.3, 12
Ex 5.3, 13
Ex 5.3, 14 Important
Ex 5.3, 15 Deleted for CBSE Board 2022 Exams
Ex 5.3, 16 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 17 Deleted for CBSE Board 2022 Exams
Ex 5.3, 18 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 19 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 20 Important Deleted for CBSE Board 2022 Exams
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Serial order wise
Transcript
Ex 5.3, 8 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. We know that an = a + (n – 1) d From (1) & (2) 14 – d = 18 – 2d Given 2nd term is 14 a2 = a + (2 – 1) d 14 = a + d 14 – d = a a = 14 – d Given 3rd term is 18 a3 = a + (3 – 1) d 18 = a + 2d 18 – 2d = a a = 18 – 2d 2d – d = 18 – 14 d = 4 Putting value of d in (1) a = 14 – d a = 14 – 4 a = 10 Now, we need to find sum of first 51 terms We can use formula Sn = 𝒏/𝟐 [2a + (n – 1) d] Putting n = 51 , a = 10 & d = 4 S51 = 51/2 (2 × 10 + (51 − 1) × 4) S51 = 51/2 (20 + 50 × 4) S51 = 51/2 × (20 + 200) S51 = 51/2 × 220 S51 = 51 × 110 S51 = 5610 Hence, the sum of first 51 terms is 5610
