Question:
The 17th term of an AP exceeds its 10th term by 21. The common difference of the AP is
(a) 3
(b) 2
(c) −3
(d) −2
Solution:
( a) 3
T10 = a + 9d
T17 = a + 16d
Also, a + 16d = 21 + T10
⇒ a + 16d = 21 + a + 9d
⇒ 7d = 21
⇒ d = 3
Hence, the common difference of the AP is 3.