Question

The 8th term of an AP is zero . Prove that it's 38th term is triple of its 18th term.

Answer 1

a8 = a +7d

a38 = 3(a18)

a+ 37d = 3(a + 17d )

a + 37d = 3a + 51d

3a - a = 51d - 17d
2a = 14
a = 14/2
a= 7

Answer 2

a8 =0
a+7d=0
a= --7d
a38=a18
a+37d=3(a+17d)
put the value of a
-7d +37d=3(-7d+17d)
30d= -21d+51d
30d =30d
hence proved
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