The largest term common to the sequences ${\rm{ }}11,{\rm{ }}21,{\rm{ }}31, \ldots \ldots \ldots $to $100$ terms find $31,35,41,45 \ldots \ldots $ to $100$ term is

The first sequence is an arithmetic progression with the first term as 11 and a common difference of 10, giving the nth term as 10n plus 1. The second sequence has a first term of 31 and alternates differences, but its terms can be represented as even positions forming one progression. Finding the common terms by inspection shows that 311 appears in both sequences. Since the question asks for the largest term within the first 100 terms of the first sequence, the last possible value is 10 times 100 plus 1, which is 1001. We find that 511 is a common term in both sequences, and any larger common term would exceed the 100-term limit or the constraints of the second sequence's differences. Therefore, the largest common term is 511.

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