Question

The 4 term of an ap is 0. Prove that the 25th term is 3 times the 11th term.

Answer 1

Given 4th term of A.P is 0
=>a+(4-1)d=0
=>a+3d=0
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let us consider given cases are equal
that is 25th term=3(11th term)
=>a+(25-1)d=3[a+(11-1)d]
=>a+24d=3a+30d
=>2a+6d=0
=>2[a+3d]=0
but a+3d=0
=>2(0)=0
=>0=0
i.e L.H.S=R.HS
=>our assumption is true
hence
25th term=3 times of 11th term..
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