Question

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

Answer 1

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Answer 2

Let the first term be a and common difference be d.
For 8th term, n=8.

Therefore, t₈=a+(n-1)d 
                  0=a+7d .....given t₈=0
                 a= -7d
Therefore, 18th term = a+17d                              38th term = a+37d
                                  = (-7d) + 17d                                      =(-7d)+37d
                                  = 10d                                                  = 30d
t₁₈/t₃₈ = 10d/30d = 1/3
∴3t₁₈=t₃₈