Question

The largest number common to both the sequences $1,11,21,31,\cdots \text{upto}100\text{terms}$ and $31,36,41,46,\cdots \text{upto}100\text{terms}$ is

• A. $531$
• B. $501$
• C. $511$
• D. $521$

Answer 1

The correct option is D $521$

Let the $n^{th}$ term of first sequence be same as $m^{th}$ term of the second sequence.
$1+(n-1)\times 10=31+(m-1)\times 5\Rightarrow 2(n-1)=6+m-1\Rightarrow 2n=m+7$
Now, maximum value of $m$ is $100$.
$2n=100+7\Rightarrow n=\frac{107}{2}$
which is not possible as $n$ is integer.

Now, $m=99$, we get
$2n=99+7=106\Rightarrow n=53$
Now, the largest number common to both of them
$=1+(53-1)\times 10=521$