If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.
a9 = 0
a + (9 – 1)d =0
a + 8d = 0
a = -8d …(i)
To Prove: a29 = 2a19
Proof: LHS= a29 = a + 28d = -8d + 28d = 20d
RHS= 2a19 = 2[a + (18)d] = 2(-8d + 18d) = 2(10d) = 20d
Since, LHS=RHS
Hence, proved