Answer:
Given,
10 times the 10th term of an A.P. is equal to 15 times the 15th term
10a10 = 15a15
an = a + (n – 1) d
When n = 10,
a10 = a + (10 – 1)d
= a + 9d
When n = 15,
a15 = a + (15 – 1)d
= a + 14d
When n = 25,
a25 = a + (25 – 1)d
= a + 24d ………(i)
10a10 = 15a15
10(a + 9d) = 15(a + 14d)
10a + 90d = 15a + 210d
10a – 15a + 90d – 210d = 0
-5a – 120d = 0
-5(a + 24d) = 0
a + 24d = 0
a25 = 0
Thus, Proved.