If (p+1)th term of an a.P.. Is twice its (q+1)th term, then (3p+1)th term
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a+(p+1-1) d=2{a+(q+1-1) d}
a+pd=2a+2qd
a+d(2q-p) =0
a=d(p-2q) -------(1)
for (3p+1) the term=dp-2qd+3qd
=dp+dq
=d(p+q)
a+pd=2a+2qd
a+d(2q-p) =0
a=d(p-2q) -------(1)
for (3p+1) the term=dp-2qd+3qd
=dp+dq
=d(p+q)
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