Question

In the series $3,7,11,15,\dots$ and $2,5,8,\dots$ each continued to $100$ terms, the number of terms that are identical is

• A. 21
• B. 27
• C. 25
• D. None of these

Answer 1

Answer: (C)

Let the $n$th term of the first series $=$ the $m$ th term of the second series.
$\therefore 3+(n-1)\times 4=2+(m-1)\times 3$,
or $4n=3m$ or $\frac{n}{3}=\frac{m}{4}=k$ (say)
$\therefore n=3k$ and $m=4k$
As each series is continued to $100$ terms,
$n=3k\le 100$ and $m=4k\le 100$
$\therefore$ Possible values of $k$ are $1,2,3,\dots ,25$ and corresponding to each value of $k$ we get one identical term.
Hence there are $25$ identical terms.