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

The formula for n^th term of an A.p is a+(n-1)d,where a is the first term,d is the common difference.

Given,4^th term=10 a.

=>10a=a+(4-1)d--(1)

=>9a=3d

=>d=3a---(2)

now,sixth term=a+(6-1)d

=>a+5d

=>a+5(3a)=16a

Thus,sixth term=4(4a)=4(2^nd term)