Given,
10 times the 10th term of an A.P. is equal to 15 times the 15th term
⇒ 10a10 = 15a15
To prove :
a25 = 0
We know,
an = a + (n – 1)d
Where a is first term or a1 and d is common difference and n is any natural number
When n = 10 :
∴ a10 = a + (10 – 1)d
⇒ a10 = a + 9d
When n = 15 :
∴ a15 = a + (15 – 1)d
⇒ a15 = a + 14d
When n = 25:
∴ a25 = a + (25 – 1)d
⇒ a25 = a + 24d ………(i)
According to question :
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 (From (i))
Hence Proved.