Ex 5.3, 12 - Chapter 5 Class 10 Arithmetic Progressions
Last updated at April 16, 2024 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
Ex 5.3, 11 Important
Ex 5.3, 12 You are here
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, 12 Find the sum of first 40 positive integers divisible by 6. Positive integers divisible by 6 are 6, 12, 18, 24,…. Since difference is same, it is an AP We need to find sum of first 40 integers We can use formula Sn = 𝑛/2 (2a + (n – 1) d) Here, n = 40 , a = 6 & d = 12 – 6 = 6 Putting values in formula Sn = 𝒏/𝟐 (2a + (n – 1) d) Sn = 40/2 (2 × 6 + (40 − 1) × 6) Sn = 20 (12 + 39 × 6) Sn = 20 (12 + 234) Sn = 20 × 246 Sn = 4920 Therefore, the sum of first 40 integers divisible by 6 is 4920
