Question

The largest term common in both the sequences $1,11,21,31,....upto\left(100\right)$ terms and $31,36,41,46,.....upto\left(100\right)$ terms is:

• A. $381$
• B. $471$
• C. $281$
• D. $521$

Answer 1

The correct option is D $521$

Given sequences $1,11,21,31,...,100terms31,36,41,46,....100terms$

These are arithmetic progressions with common difference $10,5$ respectively.

Every odd term of second AP is same with the $1^{st}$ AP. Since the number of terms for both the APs are same. The largest term is decided by the $n^{th}$ term of second AP.

Also, first AP contain all the odd terms of second AP.

The last odd term out of $100$ terms is ${99}^{th}$ term

$n^{th}$ term formula

$a_n=a+(n-1)d{a}_n=31+(99-1)5a_n=521$