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Question

The fourth term of an A.P. is ten times the first. Prove that the sixth term is four times as greater as second term.


Solution


Let first term = a & common diff = d
Fourth term = a+3d
a+3d=10a   or 3d=9a
or d=3a
2nd term = a+b = a+3a = 4a
6th term = a+5d = a+15a = 16a
6th term = 4 × 4 a 
Hence, 6th term = 4  × 2nd term

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