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Q.
In the series $3, 7, 11, 15, …$ and $2, 5, 8, …$ each continued to $100$ terms, the number of terms that are identical is
Sequences and Series
Solution:
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 $4 n=3 m$ or $\frac{n}{3}=\frac{m}{4}=k$ (say)
$\therefore n=3 k$ and $m=4 k$
As each series is continued to $100$ terms,
$n=3 k \leq 100$ and $ m=4 k \leq 100$
$\therefore $ Possible values of $k$ are $1,2,3, \ldots, 25$ and corresponding to each value of $k$ we get one identical term.
Hence there are $25$ identical terms.