Question

HELLO USERS...

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

#BE BRAINLY

Answer 1

a(8) = a+7d, where a is the first term and d is the common difference. 

a+7d=0 
a=-7d 

a(38) = a+37d = -7d+37d = 30d 
a(18)=a+17d = -7d+17d = 10d 

30 d = 3 times 10 d 

Therefore, its 38th term is triple its 18th term.

Answer 2

a+7d=0

a=-7d

a38=a+37d=-7d+37d=30d  .....(i)

a18=a+17d=-7d+17d=10d  ...(ii)

frm (i) and (ii)

hence a38=3a18

$hope \: helps$