Question

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

• A. $511$
• B. $471$
• C. $281$
• D. None of these

Answer 1

Answer: (A)

$511$

The first sequence is $1+11+21+31+41+51+61+71+81+91+...............$ upto $100th$ terms

last term will be =$1+(100-1)\times 10=991$

and for the second sequence $31+36+41+46+51+56+.................$ upto $100th$ term

last term will be=$31+(100-1)\times 5=526$

Now,

we have the common terms as $31+51+71$ and $n$ term of this sequence wil be =$31+(n-1)\times 20$

$=>20n+11$

and this will be less than $526$

So,

$=>20n+11<526$

$=>20n<526-11$

$=>n<\frac{515}{20}=25.75$

so $n=25$

Hence the largest common term is =$20\times 25\times 11=511$