Question

The number of terms common between the series $1,2,4,8,...to100$ terms and $1,4,7,10,...to100$ terms is

• A. $6$
• B. $4$
• C. $5$
• D. none of these

Answer 1

Answer: (C)

$5$

The 'mth' term in the first sequence can be written as $2^{m-1}$

The 'nth' term in the second sequence can be written as $1+(n-1)3=3n-2$

So, the numbers in the second sequence give remainder of $1$ when divided by $3$.

Now, consider the first sequence,

$2^0$ when divided by $3$ gives a remainder of $1$.

$2^1$ when divided by $3$ gives a remainder of $2$.

$2^2$ when divided by $3$ gives a remainder of $1$.

$2^3$ when divided by $3$ gives a remainder of $2$.

This cycle will follow. All even powers give a remainder of $1$ and all odd powers give remainder of $2$.....(1)

Now consider the second sequence:

The last term is $3\times 100-2=298$

For $m=10$, the term in the first sequence becomes $512$ which is more than the last term of second sequence $3n-2$.

Hence, we need to check till $m=9$.

When $m$ varies from $1$ to $9$, the powers of $2$ in the sequence $2^{m-1}$ vary from $0$ to $8$.

From (1), we need all even powers of $2$. Hence, $2^0,2^2,2^4,2^6,2^8$ are also there in the second sequence.

Hence, total $5$ numbers are common to both sequence.