Ex 5.2, 10 - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at March 19, 2021 by Teachoo
Ex 5.2
Ex 5.2, 1
Ex 5.2, 2 (i) (MCQ)
Ex 5.2, 2 (ii) (MCQ) Important
Ex 5.2, 3 (i)
Ex 5.2, 3 (ii)
Ex 5.2, 3 (iii) Important
Ex 5.2, 3 (iv)
Ex 5.2, 3 (v) Important
Ex 5.2, 4
Ex 5.2, 5 (i)
Ex 5.2, 5 (ii) Important
Ex 5.2, 6 Important
Ex 5.2, 7
Ex 5.2, 8
Ex 5.2, 9
Ex 5.2, 10 You are here
Ex 5.2, 11
Ex 5.2, 12 Important
Ex 5.2, 13 Important
Ex 5.2, 14 Important
Ex 5.2, 15
Ex 5.2, 16
Ex 5.2, 17 Important
Ex 5.2, 18
Ex 5.2, 19
Ex 5.2, 20 Important
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Serial order wise
Transcript
Ex 5.2, 10 If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference. We know that an = a + (n – 1) d So, a17 = a + (17 – 1) d a17 = a + 16d Also, a10 = a + (10 – 1)d a10 = a + 9d Given that 17th term of an A.P. exceeds its 10th term by 7 a17 – a10 = 7 (a + 16d) – (a + 9d) = 7 a + 16d – a – 9d = 7 a – a + 16d – 9d = 7 0 + 7d = 7 7d = 7 7d = 7 d = 7/7 d = 1 Therefore, common difference of the given AP is 1 (From (1) & (2))
