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Q.
The 10th common term between the series $3 + 7 + 11 + ......$ and $ 1 + 6 + 11 + ......$ is
Sequences and Series
Solution:
$T_{n} $ of $3+7+11+.....= 3+\left(n-1\right)4 = 4n-1 $
$T_{m} of 1+6+11+.....=1+\left(m-1\right)5 $
$ = 5m-4 $
For a common term, $4n - 1 = 5m - 4$
i.e., $ 4n = 5m - 3 $
For $n = 3, m = 3$ gives the first common term i.e. $11$.
$ n = 8, m = 7$ gives the second common term i.e., $31$.
$n = 13, m = 11 $ gives the third common term i.e., $51$.
$ n= 18, m = 15$ gives the fourth common term i.e., $71$.
Now, $10$th term of $3, 8, 13, 18,....$
$ = 3 + \left(10 - 1\right) 5 = 48$
For $10$ th common term, $n = 48$
$ \therefore $ reqd. common term $= 4\left(48\right)-1$
$ = 192-1 = 191$