Answer:
Given,
24th term is twice the 10th term
a24 = 2a10
an = a + (n – 1) d
When n = 10,
a10 = a + (10 – 1)d
= a + 9d
When n = 24,
a24 = a + (24 – 1)d
= a + 23d
When n = 34,
a34 = a + (34 – 1)d
= a + 33d ………(i)
When n = 72,
a72 = a + (72 – 1)d
= a + 71d
a24 = 2a10
a + 23d = 2(a + 9d)
a + 23d = 2a + 18d
a – 2a + 23d – 18d = 0
-a + 5d = 0
a = 5d
a72 = a + 71d
a72 = 5d + 71d
= 76d
= 10d + 66d
= 2(5d + 33d)
= 2(a + 33d) [since, a = 5d]
a72 = 2a34 (From (i))
Thus, Proved.