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Question

If (p+1)th term of an A.P is twice (q+1)th term, Prove that (3p+1)th term is twice the (p+q+1)th term.

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Solution

Let the first term of the A.P be a and the common difference be d
We have tp+1=2tq+1
a+(p+11)d=2[a+(q+11)d]
a+pd=2a+2qd
a=(p2q)d ..........(1)
Now, t3p+1=a+(3p+11)d
=(p2q)d+3pd=4pd2qd=2(2pq)d
t3p+1=2(2pq)d ......(2)
and tp+q+1=a+(p+q+11)d=(p2q)d+(p+q)d from (1)
tp+q+1=(2pq)d .....(3)
From (2) and (3) we have
t3p+1=2tp+q+1

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