If 8 times the 8th term of an AP is equal to 15 times the 15th term of that AP, then 23rd term of the same AP is
0
Given,
8a8=15a15
⇒ 8[a + (8 - 1)d] = 15[a + (15 - 1)d]
⇒ 8(a + 7d) = 15 (a + 14d)
⇒ 8a + 56d = 15a + 210d
⇒ -7a - 154d = 0
⇒ 7a + 154d = 0
⇒ a + 22d = 0
⇒ a + (23 - 1)d = 0