Question

The largest term common to the sequence $1,11,21,31,$ to $100$ terms and $31,36,41,\mathrm{46..}$ to $100$ terms is

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

Answer 1

Answer: (D)

$521$

$1,11,21,31,.....100$ terms

$31,36,41,46,......100$ terms

Last term of first$+$AP$=1+99\times 10=991$

Last term of second AP$=31+99\times 5=526$

$526-5=521$

$531-10=521$

Second last term of second AP$=521$

is largest common to both the sequence.

(D) is correct.