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Question

Show that a1,a2,a3,.......an form an AP where an=3+4n

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Solution

Given:-
an=3+4n
Taking n=1,
a1=3+4×1=7
a2=3+4×2=11
a3=3+4×3=15
Therefore, the series is-
7,11,15,.......
Now,
a2a1=117=4
a3a2=1511=4
Since the difference is same, i.e., 4, thus the above series form an A.P.

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