Question

The largest term common to sequence$1,11,21,31,...to100$ terms and $31,36,41,46,...to100$ terms is

• A. $381$
• B. $471$
• C. $281$
• D. none of these

Answer 1

The correct option is D none of these

Sequence 1 : 1,11,21,31,..
$t_n=1+(n-1)10=10n-9$
Sequence 2 : 31,36,41,46,...
$T_m=31+(m-1)5=26+5m$
Common term $\Rightarrow t_n=T_m\Rightarrow 10n-9=26+5m$
$\Rightarrow 10n-5m=35$
$\Rightarrow 2n=m+7$
$\therefore \forall$ odd $m,\exists$ a natural number $n$
Maximum $m$ can take is 100 here.

For $n$ to satisfy , $m$ should be odd.

$m=99$
$\Rightarrow$ Largest common term = $26+5\times 99=521$
$\therefore$ Ans. is option D.