Solve for the difference of the consecutive terms 5,8,11,14
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5,8,11,14,......
We see that the above series is in Arithmetic progression as
8−5=11−8=14−11=3
i.e. the difference between any two consecutive terms is constant through out the series.
Hence 5,8,11,14,...... is in AP, where the first term (a)=5 and common difference (d)=3
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